Work, Power & Energy
304 Questions
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Q276
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4. LINKED COMPREHENSION TYPE QUESTIONS
MCQ
Comprehension Passage: Comprehension - 21
A force $F = 50 \, \text{N}$ is applied at one end of a string, the other end of which is tied to a block of mass $10 \, \text{kg}$ . The block is free to move on a frictionless horizontal surface. Take initial instant as $\theta = 30^\circ$ and final instant as $\theta = 37^\circ$ . For the time between these two instants, answer the following questions? Take $\sin(37^\circ) = \frac{3}{5}$ .
Find the ratio of initial acceleration to final acceleration of the block.
Find the ratio of initial acceleration to final acceleration of the block.
A.
$\frac{3\sqrt{5}}{8}$
B.
$\frac{5\sqrt{3}}{8}$
C.
$\frac{8\sqrt{3}}{5}$
D.
$\frac{8\sqrt{5}}{3}$
Q277
Advance
4. LINKED COMPREHENSION TYPE QUESTIONS
MCQ
Comprehension Passage: Comprehension - 22
A particle moves in x-y plane under the action of a path dependent force $F = y\hat{i} + x\hat{j}$ . The work done by the force as it moves in x-y plane can be evaluated by solving the line integral of force i.e. $\int\vec{F}\cdot d\vec{l}$ , where $d\vec{l}=dx\hat{i}+dy\hat{j}$ . The position coordinates x and y may vary according to some constraint determined by the path followed by the particle. Based on the information shared, answer the following questions.
When the particle moves along a straight-line path from origin to the point $(a, a)$ , the work done by the force on the particle is
A.
$a^2$
B.
$2a$
C.
$\frac{a^2}{2}$
D.
Zero
Q278
Advance
4. LINKED COMPREHENSION TYPE QUESTIONS
MCQ
Comprehension Passage: Comprehension - 22
A particle moves in x-y plane under the action of a path dependent force $F = y\hat{i} + x\hat{j}$ . The work done by the force as it moves in x-y plane can be evaluated by solving the line integral of force i.e. $\int\vec{F}\cdot d\vec{l}$ , where $d\vec{l}=dx\hat{i}+dy\hat{j}$ . The position coordinates x and y may vary according to some constraint determined by the path followed by the particle. Based on the information shared, answer the following questions.
When the particle moves from $(0, 0)$ to $(a, 0)$ and then from $(a, 0)$ to $(a, a)$ in straight line paths, then work done by the force on the particle is
A.
$a^2$
B.
$2a$
C.
$\frac{a^2}{2}$
D.
Zero
Q279
Advance
4. LINKED COMPREHENSION TYPE QUESTIONS
MSQ
Comprehension Passage: Comprehension - 22
A particle moves in x-y plane under the action of a path dependent force $F = y\hat{i} + x\hat{j}$ . The work done by the force as it moves in x-y plane can be evaluated by solving the line integral of force i.e. $\int\vec{F}\cdot d\vec{l}$ , where $d\vec{l}=dx\hat{i}+dy\hat{j}$ . The position coordinates x and y may vary according to some constraint determined by the path followed by the particle. Based on the information shared, answer the following questions.
The infinitesimal work done (dW) by the given force when displaced through $d\vec{l}$ is given by
$d W = \vec {F} \cdot d \vec {l} = y d x + x d y$
From this, it can be inferred that the work done by the force
A.
depends only on initial and final values of $x$ and $y$
B.
depends on initial and final values and on the path followed
C.
is zero for any values of $x$ and $y$ .
D.
is zero when object returns to original position after following any path.
Q280
Advance
5. MATRIX MATCH/COLUMN MATCH TYPE QUESTIONS
Match the Columns
Match the following Column-I with Column-II
| Column-I | Column-II | ||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
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Correct Answer: A ā (q) B ā (s) C ā (r) D ā (r)
Explanation:
A ā (q)
$\mathrm{B} \rightarrow (\mathrm{s})$
$\mathrm{C} \rightarrow (\mathrm{r})$
(D) ā (r)
Since $W_{total} = \Delta K$ and $W_{C} = -\Delta U$
and $W_{\mathrm{ext}} = \Delta U + \Delta K = (U_f + K_f) - (U_i + K_i)$
$\mathrm{B} \rightarrow (\mathrm{s})$
$\mathrm{C} \rightarrow (\mathrm{r})$
(D) ā (r)
Since $W_{total} = \Delta K$ and $W_{C} = -\Delta U$
and $W_{\mathrm{ext}} = \Delta U + \Delta K = (U_f + K_f) - (U_i + K_i)$
Q281
Advance
5. MATRIX MATCH/COLUMN MATCH TYPE QUESTIONS
Match the Columns
A particle is suspended from a string of length $l$ and is given a velocity $u = 3\sqrt{gl}$ at the bottom. If $v_{B}, T_{B}$ and $v_{C}, T_{C}$ be the velocity, tension at points $B$ and $C$ , then match the following. Column-I with Column-II
| Column-I | Column-II | ||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
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Correct Answer: A ā (s) B ā (r) C ā (p) D ā (q)
Explanation:
A $\rightarrow$ (s)
$\mathrm{B} \rightarrow (\mathrm{r})$
$C \rightarrow (p)$
$\mathrm{D} \rightarrow (\mathrm{q})$
$v _ {B} ^ {2} = u _ {A} ^ {2} - 2 g h _ {A B} = (9 g \ell) - (2 g \ell) = 7 g \ell$
$\Rightarrow v _ {B} = \sqrt {7 g \ell}$
$\Rightarrow \frac {v _ {B} ^ {2}}{g ^ {\ell}} = 7$
Further, $T_{B}=\frac{mv_{B}^{2}}{\ell}=7mg$
$\Rightarrow \frac {T _ {B}}{2 m g} = \frac {7}{2}$
Again, $v_{C}^{2} = v_{A}^{2} - 2gh_{AC} = (9g\ell) - 2g(2\ell) = 5g\ell$
$\Rightarrow v _ {C} = \sqrt {5 g \ell}$
$\Rightarrow \frac {v _ {C} ^ {2}}{g ^ {\ell}} = 5$
Further, $T_{C} + mg = \frac{mv_{C}^{2}}{\ell}$
$\Rightarrow T _ {C} = 4 m g$
$\Rightarrow \frac {T _ {C}}{m g} = 4$
$\mathrm{B} \rightarrow (\mathrm{r})$
$C \rightarrow (p)$
$\mathrm{D} \rightarrow (\mathrm{q})$
$v _ {B} ^ {2} = u _ {A} ^ {2} - 2 g h _ {A B} = (9 g \ell) - (2 g \ell) = 7 g \ell$
$\Rightarrow v _ {B} = \sqrt {7 g \ell}$
$\Rightarrow \frac {v _ {B} ^ {2}}{g ^ {\ell}} = 7$
Further, $T_{B}=\frac{mv_{B}^{2}}{\ell}=7mg$
$\Rightarrow \frac {T _ {B}}{2 m g} = \frac {7}{2}$
Again, $v_{C}^{2} = v_{A}^{2} - 2gh_{AC} = (9g\ell) - 2g(2\ell) = 5g\ell$
$\Rightarrow v _ {C} = \sqrt {5 g \ell}$
$\Rightarrow \frac {v _ {C} ^ {2}}{g ^ {\ell}} = 5$
Further, $T_{C} + mg = \frac{mv_{C}^{2}}{\ell}$
$\Rightarrow T _ {C} = 4 m g$
$\Rightarrow \frac {T _ {C}}{m g} = 4$
Q282
Advance
5. MATRIX MATCH/COLUMN MATCH TYPE QUESTIONS
Match the Columns
A block of mass 1 kg is stationary with respect to a rough wedge making an angle of $30^{\circ}$ with horizontal as shown in figure. Starting from rest with an acceleration of $2 \, ms^{-2}$ the work done on the block at t = 4 s. Column-I with Column-II
| Column-I | Column-II | ||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
|
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Correct Answer: A ā (t) B ā (p) C ā (s) D ā (q)
Explanation:
$\mathrm{A}\rightarrow (\mathfrak{t})$
$\mathrm{B} \rightarrow (\mathrm{p})$
$\mathrm{C} \rightarrow (\mathrm{s})$
$\mathrm{D} \rightarrow (\mathrm{q})$
$\mathrm{At} t = 4 \mathrm{s},$
$v = a t = 8 \mathrm{ms} ^ {- 1}$
and $s=\frac{1}{2}at^{2}=16m$
$K E = \frac {1}{2} m v ^ {2} = 3 2 \mathrm{J}$
From Work-Energy Theorem, we have
Work done by all the forces = $\Delta KE = 32$ J
Work done by gravity
$= - m g h = - (1) (1 0) (1 6) = - 1 6 0 \mathrm{J}$
Writing equation of motion, we have, $\sum F_y = ma$
$N \cos 3 0 ^ {\circ} + f \sin 3 0 ^ {\circ} - 1 0 = m a = 2$
$\Rightarrow \sqrt {3} \mathrm{N} + f = 2 4\tag{... (1}$
$\sum F _ {x} = 0$
$\Rightarrow \quad N \sin 3 0 ^ {\circ} = f \cos 3 0 ^ {\circ}$
$\Rightarrow N = \sqrt {3} f$
... (2)
Solving equations (1) and (2), we have
$f = 6 \mathrm{N}$
and $N = \frac{18}{\sqrt{3}} = 6\sqrt{3}\mathrm{N}$
So, $W_{N} = (N\cos \theta)(s) = (6\sqrt{3})\left(\frac{\sqrt{3}}{2}\right)(16) = 144 \mathrm{~J}$
and $W_{f} = (f\sin \theta)(s) = (6)\left(\frac{1}{2}\right)(16) = 48 \, \mathrm{J}$
$\mathrm{B} \rightarrow (\mathrm{p})$
$\mathrm{C} \rightarrow (\mathrm{s})$
$\mathrm{D} \rightarrow (\mathrm{q})$
$\mathrm{At} t = 4 \mathrm{s},$
$v = a t = 8 \mathrm{ms} ^ {- 1}$
and $s=\frac{1}{2}at^{2}=16m$
$K E = \frac {1}{2} m v ^ {2} = 3 2 \mathrm{J}$
From Work-Energy Theorem, we have
Work done by all the forces = $\Delta KE = 32$ J
Work done by gravity
$= - m g h = - (1) (1 0) (1 6) = - 1 6 0 \mathrm{J}$
Writing equation of motion, we have, $\sum F_y = ma$
$N \cos 3 0 ^ {\circ} + f \sin 3 0 ^ {\circ} - 1 0 = m a = 2$
$\Rightarrow \sqrt {3} \mathrm{N} + f = 2 4\tag{... (1}$
$\sum F _ {x} = 0$
$\Rightarrow \quad N \sin 3 0 ^ {\circ} = f \cos 3 0 ^ {\circ}$
$\Rightarrow N = \sqrt {3} f$
... (2)
Solving equations (1) and (2), we have
$f = 6 \mathrm{N}$
and $N = \frac{18}{\sqrt{3}} = 6\sqrt{3}\mathrm{N}$
So, $W_{N} = (N\cos \theta)(s) = (6\sqrt{3})\left(\frac{\sqrt{3}}{2}\right)(16) = 144 \mathrm{~J}$
and $W_{f} = (f\sin \theta)(s) = (6)\left(\frac{1}{2}\right)(16) = 48 \, \mathrm{J}$
Q283
Advance
5. MATRIX MATCH/COLUMN MATCH TYPE QUESTIONS
Match the Columns
Acceleration versus x and potential energy versus x graph of a particle of mass 1 kg moving along x-axis are shown in figure. Assume that at x = 0, the velocity of the particle is $4 \, ms^{-1}$ , match the following at x = 8 m. Column-I with Column-II
| Column-I | Column-II | ||||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
|
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Correct Answer: A ā (r) B ā (q) C ā (p) D ā (t)
Explanation:
$\mathrm{A}\rightarrow (\mathrm{r})$
$\mathrm{B} \rightarrow (\mathrm{q})$
$C \rightarrow (p)$
$\mathrm{D} \rightarrow (\bar {t})$
$v _ {f} - v _ {i} = \mathrm{areaunder} a - x \mathrm{graph} = 1 2 \mathrm{ms} ^ {- 1}$
$\Rightarrow v _ {f} = 1 2 + 4 = 1 6 \mathrm{ms} ^ {- 1}$
$\Rightarrow \Delta K E = \frac {1}{2} m \left(v _ {f} ^ {2} - v _ {i} ^ {2}\right) = 1 2 0 \mathrm{J}$
Work done by all the forces = $\Delta KE = 120$ J
$K _ {f} = \frac {1}{2} m v _ {f} ^ {2} = 1 2 8 \mathrm{J}$
Work done by conservative forces = $U_{i} - U_{f} = 240$ J
Work done by external force i.e. $W_{ext}$ is given by
$W _ {\text {ext}} = \left( \begin{array}{c} \text {Total} \\ \text {work} \\ \text {done} \end{array} \right) - \left( \begin{array}{c} \text {Work done by} \\ \text {conservative} \\ \text {forces} \end{array} \right)$
$\Rightarrow W _ {\mathrm{ext}} = - 1 1 2 \mathrm{J}$
$\mathrm{B} \rightarrow (\mathrm{q})$
$C \rightarrow (p)$
$\mathrm{D} \rightarrow (\bar {t})$
$v _ {f} - v _ {i} = \mathrm{areaunder} a - x \mathrm{graph} = 1 2 \mathrm{ms} ^ {- 1}$
$\Rightarrow v _ {f} = 1 2 + 4 = 1 6 \mathrm{ms} ^ {- 1}$
$\Rightarrow \Delta K E = \frac {1}{2} m \left(v _ {f} ^ {2} - v _ {i} ^ {2}\right) = 1 2 0 \mathrm{J}$
Work done by all the forces = $\Delta KE = 120$ J
$K _ {f} = \frac {1}{2} m v _ {f} ^ {2} = 1 2 8 \mathrm{J}$
Work done by conservative forces = $U_{i} - U_{f} = 240$ J
Work done by external force i.e. $W_{ext}$ is given by
$W _ {\text {ext}} = \left( \begin{array}{c} \text {Total} \\ \text {work} \\ \text {done} \end{array} \right) - \left( \begin{array}{c} \text {Work done by} \\ \text {conservative} \\ \text {forces} \end{array} \right)$
$\Rightarrow W _ {\mathrm{ext}} = - 1 1 2 \mathrm{J}$
Q284
Advance
5. MATRIX MATCH/COLUMN MATCH TYPE QUESTIONS
Match the Columns
A pendulum is released from point A as shown in figure. At any instant, the net force on the bob is making an angle $\theta$ with the string. Then match the following.
Column-I
Column-I
| Column-I | Column-II | ||||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
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Correct Answer: A ā (p, q) B ā (t) C ā (r) D ā (s)
Explanation:
A $\rightarrow$ (p, q) B $\rightarrow$ (t) C $\rightarrow$ (r) D $\rightarrow$ (s)
Angle between net force and the string can never be obtuse. It is $90^{\circ}$ at A, $0^{\circ}$ at B and acute in between.
Angle between net force and the string can never be obtuse. It is $90^{\circ}$ at A, $0^{\circ}$ at B and acute in between.
Q285
Advance
5. MATRIX MATCH/COLUMN MATCH TYPE QUESTIONS
Match the Columns
Match the following.
Column-I
Column-I
| Column-I | Column-II | ||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
|
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Correct Answer: A ā (p, q, r) B ā (p, q, r) C ā (q, r) D ā (p, r)
Explanation:
A ā (p, q, r)
B ā (p, q, r)
C ā (q, r)
D ā (p, r)
Conceptual
B ā (p, q, r)
C ā (q, r)
D ā (p, r)
Conceptual
Q286
Advance
5. MATRIX MATCH/COLUMN MATCH TYPE QUESTIONS
Match the Columns
Match the following.
Column-I
Column-I
| Column-I | Column-II | ||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
|
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Correct Answer: A ā (r) B ā (p) C ā (q, s) D ā (q, s)
Explanation:
A ā (r)
B ā (p)
C ā (q, s)
D ā (q, s)
Conceptual
B ā (p)
C ā (q, s)
D ā (q, s)
Conceptual
Q287
Advance
5. MATRIX MATCH/COLUMN MATCH TYPE QUESTIONS
Match the Columns
Match the following.
Column-I
Column-I
| Column-I | Column-II | ||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
|
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Correct Answer: A ā (p, s) B ā (q, r) C ā (p, s) D ā (q, r)
Explanation:
A ā (p, s)
B ā (q, r)
C ā (p, s)
D ā (q, r)
Conceptual
B ā (q, r)
C ā (p, s)
D ā (q, r)
Conceptual
Q288
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5. MATRIX MATCH/COLUMN MATCH TYPE QUESTIONS
Match the Columns
A particle of mass 500 g moves along a horizontal circle of radius 16 m such that normal acceleration of particle varies with time as $a_{n} = 9t^{2}$ .
Column-I
Column-I
| Column-I | Column-II | ||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
|
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Correct Answer: A ā (s) B ā (r) C ā (p) D ā (q)
Explanation:
$\begin{array}{r l}&{\mathrm{A} \rightarrow (\mathrm{s})}\\&{\mathrm{B} \rightarrow (\mathrm{r})}\\&{\mathrm{C} \rightarrow (\mathrm{p})}\\&{\mathrm{D} \rightarrow (\mathrm{q})}\\&{\frac {v ^ {2}}{R} = 9 t ^ {2}}\\&{\Rightarrow v ^ {2} = (1 6) \big (9 t ^ {2} \big)}\\&{\Rightarrow v = 1 2 t}\\&{\Rightarrow a _ {t} = \frac {d v}{d t} = 1 2}\\&{\Rightarrow F _ {t} = m a _ {t} = \frac {1}{2} (1 2) = 6 \mathrm{N}}\\&{\mathrm{At} t = 1 \mathrm{s}, a _ {n} = \frac {v ^ {2}}{R} = 9}\\&{\Rightarrow F _ {n} = m a _ {n} = \frac {9}{2} \mathrm{N}}\\&{\Rightarrow F = \sqrt {F _ {t} ^ {2} + F _ {n} ^ {2}} = \sqrt {3 6 + \frac {8 1}{4}}}\end{array}$
$\begin{array}{r l} \Rightarrow & F = \sqrt {\frac {2 2 5}{4}} = \frac {1 5}{2} = 7. 5 \mathrm{N} \\ & P = \vec {F} \cdot \vec {v} \\ \Rightarrow & P = (F _ {t}) (v) \\ \Rightarrow & P = (6) (1 2) = 7 2 \mathrm{watt} \\ & P _ {\mathrm{av}} = \frac {W}{t} \\ \Rightarrow & P _ {\mathrm{av}} = F _ {t} (v _ {\mathrm{av}}) \end{array}$
But $v_{av}=\frac{\int\limits_{0}^{1}vdt}{\int\limits_{0}^{1}dt}=\frac{12\int\limits_{0}^{1}tdt}{1}$ $\Rightarrow \quad v_{av}=\frac{12t^{2}}{2}\bigg|_{0}^{1}=6\ ms^{-1}$ $\Rightarrow \quad P_{av}=(6)(6)=36\ watt$
$\begin{array}{r l} \Rightarrow & F = \sqrt {\frac {2 2 5}{4}} = \frac {1 5}{2} = 7. 5 \mathrm{N} \\ & P = \vec {F} \cdot \vec {v} \\ \Rightarrow & P = (F _ {t}) (v) \\ \Rightarrow & P = (6) (1 2) = 7 2 \mathrm{watt} \\ & P _ {\mathrm{av}} = \frac {W}{t} \\ \Rightarrow & P _ {\mathrm{av}} = F _ {t} (v _ {\mathrm{av}}) \end{array}$
But $v_{av}=\frac{\int\limits_{0}^{1}vdt}{\int\limits_{0}^{1}dt}=\frac{12\int\limits_{0}^{1}tdt}{1}$ $\Rightarrow \quad v_{av}=\frac{12t^{2}}{2}\bigg|_{0}^{1}=6\ ms^{-1}$ $\Rightarrow \quad P_{av}=(6)(6)=36\ watt$
Q289
Advance
5. MATRIX MATCH/COLUMN MATCH TYPE QUESTIONS
Match the Columns
A bob of mass 2 kg is suspended from a vehicle by a rope of length l = 5 m. The vehicle and the bob are moving at a constant speed $v_{0}$ . The vehicle is suddenly stopped by a bumper and the bob on the rope swings out a maximum angle of $60^{\circ}$ . Taking $g = 10 \, ms^{-2}$ , match the following.
Column-I
Column-I
| Column-I | Column-II | ||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
|
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Correct Answer: A ā (s) B ā (r) C ā (q) D ā (p)
Explanation:
A ā (s)
B ā (r)
C ā (q)
D ā (p)
Just after the vehicle is stopped, the bob swings out to an angle of $60^{\circ}$ . So, by Law of Conservation of Mechanical Energy, we have
$\begin{array}{r l} & m g h = \frac {1}{2} m v _ {0} ^ {2} \\ \Rightarrow & \frac {1}{2} m v _ {0} ^ {2} = m g \ell (1 - \cos \theta) \\ \Rightarrow & v _ {0} = \sqrt {2 (1 0) (5) \left(1 - \frac {1}{2}\right)} = 5 \sqrt {2} \mathrm{ms} ^ {- 1} \end{array}$
So, net force on the bob at the lowest point is
$F = \frac {m v _ {0} ^ {2}}{\ell} = \frac {(2) (5 0)}{5} = 2 0 \mathrm{N}$
Acceleration of the bob at lowest point is
$a = \frac {F}{m} = \frac {v _ {0} ^ {2}}{\ell} = 1 0 \mathrm{ms} ^ {- 2}$
At the highest point, net force is
$F = m g \sin (6 0 ^ {\circ}) = 1 0 \sqrt {3} \mathrm{N}$
So, acceleration at highest point is
$a = \frac {F}{m} = 5 \sqrt {3} \mathrm{ms} ^ {- 2}$
B ā (r)
C ā (q)
D ā (p)
Just after the vehicle is stopped, the bob swings out to an angle of $60^{\circ}$ . So, by Law of Conservation of Mechanical Energy, we have
$\begin{array}{r l} & m g h = \frac {1}{2} m v _ {0} ^ {2} \\ \Rightarrow & \frac {1}{2} m v _ {0} ^ {2} = m g \ell (1 - \cos \theta) \\ \Rightarrow & v _ {0} = \sqrt {2 (1 0) (5) \left(1 - \frac {1}{2}\right)} = 5 \sqrt {2} \mathrm{ms} ^ {- 1} \end{array}$
So, net force on the bob at the lowest point is
$F = \frac {m v _ {0} ^ {2}}{\ell} = \frac {(2) (5 0)}{5} = 2 0 \mathrm{N}$
Acceleration of the bob at lowest point is
$a = \frac {F}{m} = \frac {v _ {0} ^ {2}}{\ell} = 1 0 \mathrm{ms} ^ {- 2}$
At the highest point, net force is
$F = m g \sin (6 0 ^ {\circ}) = 1 0 \sqrt {3} \mathrm{N}$
So, acceleration at highest point is
$a = \frac {F}{m} = 5 \sqrt {3} \mathrm{ms} ^ {- 2}$
Q290
Advance
6. INTEGER/NUMERICAL ANSWER TYPE QUESTIONS
Numerical
A small block of mass 1 kg is kept on a rough inclined wedge of inclination $45^{\circ}$ fixed in an elevator. The elevator goes up with a uniform velocity $v = 2 \, ms^{-1}$ and the block does not slide on the wedge. Calculate the work done, in joule by the force of friction on the block in 1 s. Take $g = 10 \, ms^{-2}$ .
Correct Answer: 10
Explanation:
Since the block does not slide on the wedge. So
$f = m g \sin (4 5 ^ {\circ}) = \frac {(1) (1 0)}{\sqrt {2}} = \frac {1 0}{\sqrt {2}} \mathrm{N}$
$\begin{array}{r l} & {\mathrm{Now} W _ {f} = \vec {f} \cdot \vec {s}} \\ & {\Rightarrow W _ {f} = (f \cos 4 5 ^ {\circ}) (2)} \\ & {\Rightarrow W _ {f} = \bigg (\frac {1 0}{\sqrt {2}} \bigg) \bigg (\frac {1}{\sqrt {2}} \bigg) (2)} \\ & {\Rightarrow W _ {f} = 1 0 \mathrm{J}} \end{array}$
$f = m g \sin (4 5 ^ {\circ}) = \frac {(1) (1 0)}{\sqrt {2}} = \frac {1 0}{\sqrt {2}} \mathrm{N}$
$\begin{array}{r l} & {\mathrm{Now} W _ {f} = \vec {f} \cdot \vec {s}} \\ & {\Rightarrow W _ {f} = (f \cos 4 5 ^ {\circ}) (2)} \\ & {\Rightarrow W _ {f} = \bigg (\frac {1 0}{\sqrt {2}} \bigg) \bigg (\frac {1}{\sqrt {2}} \bigg) (2)} \\ & {\Rightarrow W _ {f} = 1 0 \mathrm{J}} \end{array}$
Q291
Advance
6. INTEGER/NUMERICAL ANSWER TYPE QUESTIONS
Numerical
A train of mass $2 \times 10^{5}$ kg has a constant speed of $20 \, ms^{-1}$ up a hill inclined at $\theta = \sin^{-1}\left(\frac{1}{50}\right)$ to the horizontal when the engine is working at $8 \times 10^{5}$ watt. Calculate the resistance to the motion of the train, in newton. Take $g = 9.8 \, ms^{-2}$ .
Correct Answer: 800
Explanation:
Since, P = Fv
$\Rightarrow F = \frac {P}{v} = \frac {8 \times 1 0 ^ {5}}{2 0} = 4 \times 1 0 ^ {4} \mathrm{N}$
At constant speed, the forces acting on the train are in equilibrium. Resolving the forces parallel to the hill
$\begin{array}{r l} & F = R + (2 \times 1 0 ^ {5}) \times g \times \left(\frac {1}{5 0}\right) \\ \Rightarrow & 4 \times 1 0 ^ {4} = R + 3 9 2 0 0 \\ \Rightarrow & R = 8 0 0 \mathrm{N} \end{array}$
Therefore, the resistance is 800 N.
$\Rightarrow F = \frac {P}{v} = \frac {8 \times 1 0 ^ {5}}{2 0} = 4 \times 1 0 ^ {4} \mathrm{N}$
At constant speed, the forces acting on the train are in equilibrium. Resolving the forces parallel to the hill
$\begin{array}{r l} & F = R + (2 \times 1 0 ^ {5}) \times g \times \left(\frac {1}{5 0}\right) \\ \Rightarrow & 4 \times 1 0 ^ {4} = R + 3 9 2 0 0 \\ \Rightarrow & R = 8 0 0 \mathrm{N} \end{array}$
Therefore, the resistance is 800 N.
Q292
Advance
6. INTEGER/NUMERICAL ANSWER TYPE QUESTIONS
Numerical
A block of mass m = 4 kg is dragged 2 m along a horizontal surface by a force F = 30 N acting at $53^{\circ}$ to the horizontal. The initial speed is $3 \, ms^{-1}$ and $\mu_{k} = \frac{1}{8}$ . Given that $\cos(37^{\circ}) = \frac{3}{5}$ .
(A) Find the change in kinetic energy of the block, in joule.
(B) Find its final speed, in $\mathrm{cms}^{-1}$ .
(A) Find the change in kinetic energy of the block, in joule.
(B) Find its final speed, in $\mathrm{cms}^{-1}$ .
Correct Answer: (a) 32, (b) 500
Explanation:
(a) The forces acting on the block are shown in the figure.
Clearly, $W_{N}=0$ and $W_{g}=0$ , whereas $W_{F}=Fs\cos\theta$
$W _ {f} = - f s = - \mu_ {k} N s \text { where } N = m g - F \sin \theta$
According to Work Energy Theorem,
$\begin{array}{r l} & {\Delta K = W _ {n e t} = W _ {F} + W _ {f}} \\ {\Rightarrow} & {\Delta K = F s \mathrm{cos} \theta - \mu_ {\mathrm{k}} (m g - F \mathrm{sin} \theta) s} \\ {\Rightarrow} & {\Delta K = (3 0) (2) (0. 6) - \frac {1}{8} (4 0 - 2 4) (2) = 3 2 \mathrm{J}} \end{array}$
b) Now $\Delta K=\frac{1}{2}mv_{f}^{2}-\frac{1}{2}mv_{i}^{2}=32$ J
Since $v_{i}=3\ ms^{-1}$
therefore, $v_{f} = 5 \, ms^{-1} = 500 \, cms^{-1}$
Clearly, $W_{N}=0$ and $W_{g}=0$ , whereas $W_{F}=Fs\cos\theta$
$W _ {f} = - f s = - \mu_ {k} N s \text { where } N = m g - F \sin \theta$
According to Work Energy Theorem,
$\begin{array}{r l} & {\Delta K = W _ {n e t} = W _ {F} + W _ {f}} \\ {\Rightarrow} & {\Delta K = F s \mathrm{cos} \theta - \mu_ {\mathrm{k}} (m g - F \mathrm{sin} \theta) s} \\ {\Rightarrow} & {\Delta K = (3 0) (2) (0. 6) - \frac {1}{8} (4 0 - 2 4) (2) = 3 2 \mathrm{J}} \end{array}$
b) Now $\Delta K=\frac{1}{2}mv_{f}^{2}-\frac{1}{2}mv_{i}^{2}=32$ J
Since $v_{i}=3\ ms^{-1}$
therefore, $v_{f} = 5 \, ms^{-1} = 500 \, cms^{-1}$
Q293
Advance
6. INTEGER/NUMERICAL ANSWER TYPE QUESTIONS
Numerical
A small block of mass $m$ slides along a smooth frictionless track as shown in the figure.
(A) If it starts from rest at P, the resultant force acting on it at Q is $\sqrt{x}$ mg. Calculate x
(B) At height $h = *R$ above the bottom of the loop should the block be released so that the force it exerts against the track at the top of the loop equals its weight. Find* .
(A) If it starts from rest at P, the resultant force acting on it at Q is $\sqrt{x}$ mg. Calculate x
(B) At height $h = *R$ above the bottom of the loop should the block be released so that the force it exerts against the track at the top of the loop equals its weight. Find* .
Correct Answer: (a) 65, (b) 3
Explanation:
(a) By Law of Conservation of Energy, we get
$\begin{array}{r l} & 0 + m g (4 R) = \frac {1}{2} m v ^ {2} \\ \Rightarrow & 4 g R = \frac {v ^ {2}}{2} \end{array}$
At $Q$ , we have
$N = \frac {m v ^ {2}}{R} = \frac {m (8 g R)}{R} = 8 m g$
So, net force at $Q$ is
$\begin{array}{r l} & F _ {Q} = \sqrt {N ^ {2} + (m g) ^ {2}} \\ \Rightarrow & F _ {Q} = \sqrt {6 4 m ^ {2} g ^ {2} + m ^ {2} g ^ {2}} = \sqrt {6 5} m g \end{array}$
$\Rightarrow x = 6 5$
b) For block to exert force (on the top of track) equal to weight, we have
N = m g
Since $mg + N = \frac{mv^{2}}{R}$
$\Rightarrow \frac {m v ^ {2}}{R} = 2 m g$
$\Rightarrow v ^ {2} = 2 R g$
Applying Law of Conservation of Energy, we get
$\begin{array}{r l} & m g (h) = \frac {1}{2} m (2 R g) + m g (2 R) \\ \Rightarrow & h = 3 R \\ \Rightarrow & * = 3 \end{array}$
$\begin{array}{r l} & 0 + m g (4 R) = \frac {1}{2} m v ^ {2} \\ \Rightarrow & 4 g R = \frac {v ^ {2}}{2} \end{array}$
At $Q$ , we have
$N = \frac {m v ^ {2}}{R} = \frac {m (8 g R)}{R} = 8 m g$
So, net force at $Q$ is
$\begin{array}{r l} & F _ {Q} = \sqrt {N ^ {2} + (m g) ^ {2}} \\ \Rightarrow & F _ {Q} = \sqrt {6 4 m ^ {2} g ^ {2} + m ^ {2} g ^ {2}} = \sqrt {6 5} m g \end{array}$
$\Rightarrow x = 6 5$
b) For block to exert force (on the top of track) equal to weight, we have
N = m g
Since $mg + N = \frac{mv^{2}}{R}$
$\Rightarrow \frac {m v ^ {2}}{R} = 2 m g$
$\Rightarrow v ^ {2} = 2 R g$
Applying Law of Conservation of Energy, we get
$\begin{array}{r l} & m g (h) = \frac {1}{2} m (2 R g) + m g (2 R) \\ \Rightarrow & h = 3 R \\ \Rightarrow & * = 3 \end{array}$
Q294
Advance
6. INTEGER/NUMERICAL ANSWER TYPE QUESTIONS
Numerical
A body slides down an inclined surface which ends into a vertical loop of radius R = 40 cm as shown in figure.
What must be the height H, in cm, of the inclined surface for the body not to fall at the uppermost point of the loop? Assume friction to be absent.
What must be the height H, in cm, of the inclined surface for the body not to fall at the uppermost point of the loop? Assume friction to be absent.
Correct Answer: 100
Explanation:
Let v be the velocity of the particle at the highest point.
According to Newton's Second Law, the net force
$F_{net} = N + mg$ provides the centripetal force to the body to revolve in a circle of radius R. So
$N + m g = \frac {m v ^ {2}}{R}$
The body does not fall at the uppermost point of the loop if $N \geq 0$ . In the limiting case, N = 0. So,
$\begin{array}{r l} & m g = \frac {m v ^ {2}}{R} \\ \Rightarrow & v ^ {2} = g R \end{array}\tag{... (1}$
Applying Law of Conservation of Energy at the initial and highest point of the loop, we get
$m g H = m g (2 R) + \frac {1}{2} m v ^ {2}$
From (1), $v^{2} = gR$ , so, we get
$m g H = m g (2 R) + \frac {1}{2} m (g R)$
$\Rightarrow H = \frac {5}{2} R = 2. 5 R$
Since $R = 40 \, \mathrm{cm}$ , we get
$\Rightarrow H = (2. 5) (4 0) = 1 0 0 \mathrm{cm}$
According to Newton's Second Law, the net force
$F_{net} = N + mg$ provides the centripetal force to the body to revolve in a circle of radius R. So
$N + m g = \frac {m v ^ {2}}{R}$
The body does not fall at the uppermost point of the loop if $N \geq 0$ . In the limiting case, N = 0. So,
$\begin{array}{r l} & m g = \frac {m v ^ {2}}{R} \\ \Rightarrow & v ^ {2} = g R \end{array}\tag{... (1}$
Applying Law of Conservation of Energy at the initial and highest point of the loop, we get
$m g H = m g (2 R) + \frac {1}{2} m v ^ {2}$
From (1), $v^{2} = gR$ , so, we get
$m g H = m g (2 R) + \frac {1}{2} m (g R)$
$\Rightarrow H = \frac {5}{2} R = 2. 5 R$
Since $R = 40 \, \mathrm{cm}$ , we get
$\Rightarrow H = (2. 5) (4 0) = 1 0 0 \mathrm{cm}$
Q295
Advance
6. INTEGER/NUMERICAL ANSWER TYPE QUESTIONS
Numerical
A block of mass $m = 2 \, \mathrm{kg}$ is attached to a spring whose spring constant is $k = 8 \, \mathrm{Nm}^{-1}$ . The block slides on an incline for which $\mu_k = \frac{1}{8}$ and $\theta = 37^\circ$ as shown in figure.
If the block starts at rest with the spring unextended, what is its speed, in $ms^{-1}$ , when it has moved a distance d = 0.5 m down the incline?
If the block starts at rest with the spring unextended, what is its speed, in $ms^{-1}$ , when it has moved a distance d = 0.5 m down the incline?
Correct Answer: 2
Explanation:
Since $W_{nc} + W_{ext} = \Delta U + \Delta K$
Let the block move down the incline through a distance x, then
$\begin{array}{r l} & f x \cos (1 8 0 ^ {\circ}) + 0 = \frac {1}{2} k x ^ {2} + (- m g x \sin \theta) + \frac {1}{2} m v ^ {2} \\ \Rightarrow & - (\mu_ {k} m g \cos \theta) x = \frac {1}{2} k x ^ {2} - m g x \sin \theta + \frac {1}{2} m v ^ {2} \end{array}$
$\Rightarrow - 1 = 1 - 6 + v ^ {2}$
$\Rightarrow v ^ {2} = 4$
$\Rightarrow v = 2 \mathrm{ms} ^ {- 1}$
Let the block move down the incline through a distance x, then
$\begin{array}{r l} & f x \cos (1 8 0 ^ {\circ}) + 0 = \frac {1}{2} k x ^ {2} + (- m g x \sin \theta) + \frac {1}{2} m v ^ {2} \\ \Rightarrow & - (\mu_ {k} m g \cos \theta) x = \frac {1}{2} k x ^ {2} - m g x \sin \theta + \frac {1}{2} m v ^ {2} \end{array}$
$\Rightarrow - 1 = 1 - 6 + v ^ {2}$
$\Rightarrow v ^ {2} = 4$
$\Rightarrow v = 2 \mathrm{ms} ^ {- 1}$
Q296
Advance
6. INTEGER/NUMERICAL ANSWER TYPE QUESTIONS
Numerical
A small block of ice with mass 120 g is placed against a horizontal compressed spring mounted on a horizontal table top that is 2.5 m above the floor as shown in figure.
The spring has a force constant $k = 2400 \, Nm^{-1}$ and is initially compressed 5 cm. The mass of the spring is negligible. The spring is released and the block slides along the table, goes off the edge and travels to the floor. If there is negligible, friction between the ice and the table and ice does not melt in the process, what is the speed of the block of ice, in $ms^{-1}$ , when it reaches the floor. Take $g = 10 \, ms^{-2}$
The spring has a force constant $k = 2400 \, Nm^{-1}$ and is initially compressed 5 cm. The mass of the spring is negligible. The spring is released and the block slides along the table, goes off the edge and travels to the floor. If there is negligible, friction between the ice and the table and ice does not melt in the process, what is the speed of the block of ice, in $ms^{-1}$ , when it reaches the floor. Take $g = 10 \, ms^{-2}$
Correct Answer: 10
Explanation:
Let u be the speed of ice block when it leaves the table top. Then
$\begin{array}{r l} & {\frac {1}{2} m u ^ {2} = \frac {1}{2} k x ^ {2}} \\ {\Rightarrow} & {u ^ {2} = \left(\frac {k}{m}\right) x ^ {2}} \end{array}$
If v be its speed when it reaches the floor, then By Law of Conservation of Energy, we have
$\begin{array}{r l} & (U + K) _ {\mathrm{atA}} = (U + K) _ {\mathrm{atB}} \\ \Rightarrow & \frac {1}{2} m u ^ {2} + m g h = \frac {1}{2} m v ^ {2} + 0 \\ \Rightarrow & v = \sqrt {u ^ {2} + 2 g h} = \sqrt {\frac {k x ^ {2}}{m} + 2 g h} \\ \text {Now,} & x = 5 \mathrm{cm} = 0.0 5 \mathrm{m} \\ & k = 2 4 0 0 \mathrm{Nm} ^ {- 1} \\ & m = 1 2 0 \mathrm{g} = 0.1 2 \mathrm{kg} \\ & h = 2. 5 \mathrm{m} \\ \Rightarrow & v = \sqrt {\frac {(2 4 0 0) (0.0 5) ^ {2}}{0.1 2} + 2 (1 0) (2.5)} \\ \Rightarrow & v = \sqrt {5 0 + 5 0} = 1 0 \mathrm{ms} ^ {- 1} \end{array}$
$\begin{array}{r l} & {\frac {1}{2} m u ^ {2} = \frac {1}{2} k x ^ {2}} \\ {\Rightarrow} & {u ^ {2} = \left(\frac {k}{m}\right) x ^ {2}} \end{array}$
If v be its speed when it reaches the floor, then By Law of Conservation of Energy, we have
$\begin{array}{r l} & (U + K) _ {\mathrm{atA}} = (U + K) _ {\mathrm{atB}} \\ \Rightarrow & \frac {1}{2} m u ^ {2} + m g h = \frac {1}{2} m v ^ {2} + 0 \\ \Rightarrow & v = \sqrt {u ^ {2} + 2 g h} = \sqrt {\frac {k x ^ {2}}{m} + 2 g h} \\ \text {Now,} & x = 5 \mathrm{cm} = 0.0 5 \mathrm{m} \\ & k = 2 4 0 0 \mathrm{Nm} ^ {- 1} \\ & m = 1 2 0 \mathrm{g} = 0.1 2 \mathrm{kg} \\ & h = 2. 5 \mathrm{m} \\ \Rightarrow & v = \sqrt {\frac {(2 4 0 0) (0.0 5) ^ {2}}{0.1 2} + 2 (1 0) (2.5)} \\ \Rightarrow & v = \sqrt {5 0 + 5 0} = 1 0 \mathrm{ms} ^ {- 1} \end{array}$
Q297
Advance
6. INTEGER/NUMERICAL ANSWER TYPE QUESTIONS
Numerical
Figure shows, a 4 kg block accelerated by a compressed spring whose spring constant is $640 \, Nm^{-1}$ . After leaving the spring at the spring's relaxed length, the block travels over a horizontal surface, with a coefficient of kinetic friction of 0.25, for a distance of 8 m before stopping.
(A) Calculate the increase in the thermal energy, in joule, of the block-floor system.
(B) Calculate the maximum kinetic energy of the block, in joule.
(C) Through what distance is the spring compressed, in cm, before the block begins to move?
(A) Calculate the increase in the thermal energy, in joule, of the block-floor system.
(B) Calculate the maximum kinetic energy of the block, in joule.
(C) Through what distance is the spring compressed, in cm, before the block begins to move?
Correct Answer: (a) 80, (b) 80, (c) 50
Explanation:
$(a) \text { Increase in thermal energy of block - floor system is }$
the work done against friction.
$\Rightarrow W _ {\text { friction }} = \mu_ {K} m g s = (0.2 5) (4) (1 0) (8)$
$\Rightarrow W _ {\text { friction }} \approx 8 0 \mathrm{J}$
b) Maximum kinetic energy of the block is the mechanical energy of spring block system which actually is the energy dissipated as heat = 80 J
c) $80 = \frac{1}{2} kx^{2} = \frac{1}{2}(640)x^{2}$
$\Rightarrow$ x = 0.5 m = 50 cm
the work done against friction.
$\Rightarrow W _ {\text { friction }} = \mu_ {K} m g s = (0.2 5) (4) (1 0) (8)$
$\Rightarrow W _ {\text { friction }} \approx 8 0 \mathrm{J}$
b) Maximum kinetic energy of the block is the mechanical energy of spring block system which actually is the energy dissipated as heat = 80 J
c) $80 = \frac{1}{2} kx^{2} = \frac{1}{2}(640)x^{2}$
$\Rightarrow$ x = 0.5 m = 50 cm
Q298
Advance
6. INTEGER/NUMERICAL ANSWER TYPE QUESTIONS
Numerical
It is observed that when the 2 kg smooth collar is slid back to point B and released from rest, the collar attains a speed of $2 \, ms^{-1}$ when it arrives at point A as shown in figure.
Calculate the necessary unstretched length of the spring, in metre, if it has a stiffness $k = 2 \, Nm^{-1}$ .
Calculate the necessary unstretched length of the spring, in metre, if it has a stiffness $k = 2 \, Nm^{-1}$ .
Correct Answer: 3
Explanation:
Let $\ell_{0}$ be the unstretched length of the spring, then by Law of Conservation of Energy, we have
$\begin{array}{r l} & (U + K) _ {\mathrm{atA}} = (U + K) _ {\mathrm{atB}} \\ & \frac {1}{2} k (3 - \ell_ {0}) ^ {2} + \frac {1}{2} m v _ {A} ^ {2} = \frac {1}{2} k (5 - \ell_ {0}) ^ {2} + \frac {1}{2} m v _ {B} ^ {2} \\ \text {But} v _ {B} = 0 \\ \Rightarrow & (3 - \ell_ {0}) ^ {2} + 4 = (5 - \ell_ {0}) ^ {2} \\ \Rightarrow & (9 + \not \ell_ {0} ^ {2} - 6 \ell_ {0}) + 4 = (2 5 + \not \ell_ {0} ^ {2} - 1 0 \ell_ {0}) \\ \Rightarrow & 1 6 - 4 \ell_ {0} = 4 \\ \Rightarrow & \ell_ {0} = 3 \mathrm{m} \end{array}$
$\begin{array}{r l} & (U + K) _ {\mathrm{atA}} = (U + K) _ {\mathrm{atB}} \\ & \frac {1}{2} k (3 - \ell_ {0}) ^ {2} + \frac {1}{2} m v _ {A} ^ {2} = \frac {1}{2} k (5 - \ell_ {0}) ^ {2} + \frac {1}{2} m v _ {B} ^ {2} \\ \text {But} v _ {B} = 0 \\ \Rightarrow & (3 - \ell_ {0}) ^ {2} + 4 = (5 - \ell_ {0}) ^ {2} \\ \Rightarrow & (9 + \not \ell_ {0} ^ {2} - 6 \ell_ {0}) + 4 = (2 5 + \not \ell_ {0} ^ {2} - 1 0 \ell_ {0}) \\ \Rightarrow & 1 6 - 4 \ell_ {0} = 4 \\ \Rightarrow & \ell_ {0} = 3 \mathrm{m} \end{array}$
Q299
Advance
6. INTEGER/NUMERICAL ANSWER TYPE QUESTIONS
Numerical
Block A has a weight of 300 N and block B has a weight of 50 N. If the coefficient of kinetic friction between the incline and block A is $\mu_{k}=\frac{1}{5}$ as shown in figure.
The speed, in ms $^{-1}$ , of block A after it moves 1 m down the plane, starting from rest is $\frac{4}{5}\sqrt{*}$ , where* is not readable. Find*. Neglect the mass of the cord and pulleys. Take $g = 10 \, ms^{-2}$ .
The speed, in ms $^{-1}$ , of block A after it moves 1 m down the plane, starting from rest is $\frac{4}{5}\sqrt{*}$ , where* is not readable. Find*. Neglect the mass of the cord and pulleys. Take $g = 10 \, ms^{-2}$ .
Correct Answer: 2
Explanation:
Using our knowledge of constraint relations, we observe that at any instant $x_{B}=2x_{A}$ and $v_{B}=2v_{A}$ . So, let $v_{A}=v$ after $x_{A}=1m$ . Using Work Energy Theorem for a non-conservative system, we get $W_{nc}=\Delta U+\Delta K$ .
Before we make use of the above relation, we must observe that
(i) A move down by $x_{A}$ , so its potential energy decreases by $m_{A}gh_{A}$ where $h_{A} = x_{A} \sin \theta = (1)\left(\frac{3}{5}\right) = \frac{3}{5} \, \text{m}.$
(ii) B moves up by $x_{B} = h_{B}$ , so its potential energy as well as kinetic energy increases.
$\Rightarrow f_{K}x_{A}\cos (180^{\circ}) = \Delta U_{A} + \Delta K_{A} + \Delta U_{B} + \Delta K_{B}$ where $f_{K} = \mu_{K}m_{A}g\cos \theta ,\quad \cos \theta = \frac{4}{5},\quad x_{A} = 1\mathrm{m},$ $x_{B} = 2\mathrm{m},\quad h_{A} = 0.6\mathrm{m},\quad m_{A} = 30\mathrm{kg},\quad m_{B} = 5\mathrm{kg},$ $g = 10~\mathrm{ms^{-2}}$
$\begin{array}{r l} \Rightarrow & - (\mu_ {K} m _ {A} g \cos \theta) x _ {A} = - m _ {A} g h _ {A} + m _ {B} g h _ {B} + \\ & \frac {1}{2} m _ {A} v ^ {2} + \frac {1}{2} m _ {B} (2 v) ^ {2} \end{array}$
Substituting the values, we get
$\begin{array}{r l} & {- \bigg (\frac {1}{5} \bigg) (3 0 0) \bigg (\frac {4}{5} \bigg) = - (3 0 0) \bigg (\frac {3}{5} \bigg) + (5 0) (2) +} \\ & {\qquad \qquad \qquad \qquad \qquad \frac {1}{2} (3 0) v ^ {2} + \frac {1}{2} (5) (2 v) ^ {2}} \\ {\Rightarrow} & {- 4 8 = - 1 8 0 + 1 0 0 + 2 5 v ^ {2}} \\ {\Rightarrow} & {v = \sqrt {\frac {3 2}{2 5}} = \frac {4 \sqrt {2}}{5}} \\ {\Rightarrow} & {* = 2} \end{array}$
Before we make use of the above relation, we must observe that
(i) A move down by $x_{A}$ , so its potential energy decreases by $m_{A}gh_{A}$ where $h_{A} = x_{A} \sin \theta = (1)\left(\frac{3}{5}\right) = \frac{3}{5} \, \text{m}.$
(ii) B moves up by $x_{B} = h_{B}$ , so its potential energy as well as kinetic energy increases.
$\Rightarrow f_{K}x_{A}\cos (180^{\circ}) = \Delta U_{A} + \Delta K_{A} + \Delta U_{B} + \Delta K_{B}$ where $f_{K} = \mu_{K}m_{A}g\cos \theta ,\quad \cos \theta = \frac{4}{5},\quad x_{A} = 1\mathrm{m},$ $x_{B} = 2\mathrm{m},\quad h_{A} = 0.6\mathrm{m},\quad m_{A} = 30\mathrm{kg},\quad m_{B} = 5\mathrm{kg},$ $g = 10~\mathrm{ms^{-2}}$
$\begin{array}{r l} \Rightarrow & - (\mu_ {K} m _ {A} g \cos \theta) x _ {A} = - m _ {A} g h _ {A} + m _ {B} g h _ {B} + \\ & \frac {1}{2} m _ {A} v ^ {2} + \frac {1}{2} m _ {B} (2 v) ^ {2} \end{array}$
Substituting the values, we get
$\begin{array}{r l} & {- \bigg (\frac {1}{5} \bigg) (3 0 0) \bigg (\frac {4}{5} \bigg) = - (3 0 0) \bigg (\frac {3}{5} \bigg) + (5 0) (2) +} \\ & {\qquad \qquad \qquad \qquad \qquad \frac {1}{2} (3 0) v ^ {2} + \frac {1}{2} (5) (2 v) ^ {2}} \\ {\Rightarrow} & {- 4 8 = - 1 8 0 + 1 0 0 + 2 5 v ^ {2}} \\ {\Rightarrow} & {v = \sqrt {\frac {3 2}{2 5}} = \frac {4 \sqrt {2}}{5}} \\ {\Rightarrow} & {* = 2} \end{array}$
Q300
Advance
6. INTEGER/NUMERICAL ANSWER TYPE QUESTIONS
Numerical
The sphere at A is given a downward velocity $v_{0}$ of magnitude $5 \, ms^{-1}$ and swings in a vertical plane at the end of a rope of length l = 2 m attached to a support at O as shown in figure.
The angle at which the rope will break is $\theta = \sin^{-1}\left(\frac{1}{*}\right)$ , where * is not readable knowing that the maximum tension that the rope can withstand is twice the weight of the sphere and taking $g = 10 \, ms^{-2}$ , find*.
The angle at which the rope will break is $\theta = \sin^{-1}\left(\frac{1}{*}\right)$ , where * is not readable knowing that the maximum tension that the rope can withstand is twice the weight of the sphere and taking $g = 10 \, ms^{-2}$ , find*.
Correct Answer: 4
Explanation:
$h = \ell \sin \theta = 2 \sin \theta$
$\begin{array}{r l} & v ^ {2} = v _ {0} ^ {2} + 2 g h = 2 5 + 2 \times 1 0 \times 2 \sin \theta \\ & \Rightarrow v ^ {2} = 2 5 + 4 0 \sin \theta \\ & \text {Further,} T - m g \sin \theta = \frac {m v ^ {2}}{\ell} \\ & \text {Given,} T = 2 m g \\ & \Rightarrow 2 m g - m g \sin \theta = \frac {m v ^ {2}}{\ell} \\ & \Rightarrow 2 (1 0) - 1 0 \sin \theta = \frac {2 5 + 4 0 \sin \theta}{2} \\ & \Rightarrow \sin \theta = \frac {1}{4} \\ & \Rightarrow \theta = \sin^ {- 1} \left(\frac {1}{4}\right) \\ & \text {So,} * = 4 \end{array}$
$\begin{array}{r l} & v ^ {2} = v _ {0} ^ {2} + 2 g h = 2 5 + 2 \times 1 0 \times 2 \sin \theta \\ & \Rightarrow v ^ {2} = 2 5 + 4 0 \sin \theta \\ & \text {Further,} T - m g \sin \theta = \frac {m v ^ {2}}{\ell} \\ & \text {Given,} T = 2 m g \\ & \Rightarrow 2 m g - m g \sin \theta = \frac {m v ^ {2}}{\ell} \\ & \Rightarrow 2 (1 0) - 1 0 \sin \theta = \frac {2 5 + 4 0 \sin \theta}{2} \\ & \Rightarrow \sin \theta = \frac {1}{4} \\ & \Rightarrow \theta = \sin^ {- 1} \left(\frac {1}{4}\right) \\ & \text {So,} * = 4 \end{array}$