Work, Power & Energy
227 Questions
Start Allen Test
Q151
Allen
2. Dynamics of Circular Motion
Numerical
For traffic moving at 60 km/hr, if the radius of the curve is 0.1 km, what is the correct angle of banking of the road? $(g = 10 \, m/s^{2})$
Correct Answer: $\theta = tan^{-1}\left(\frac{5}{18}\right)$
Explanation:
Ans. $\theta = tan^{-1}\left(\frac{5}{18}\right)$
In case of banking $\tan \theta = \frac{v^2}{rg}$ Here $v = 60km / hr = 60\times \frac{5}{18} ms - 1$ $= \frac{50}{3} ms^{-1}r = 0.1km = 100m$ So $\tan \theta = \frac{\left(\frac{50}{3}\right)^2}{100\times 10} = \frac{5}{18}\quad \Rightarrow \quad \theta = \tan^{-1}\left(\frac{5}{18}\right)$
In case of banking $\tan \theta = \frac{v^2}{rg}$ Here $v = 60km / hr = 60\times \frac{5}{18} ms - 1$ $= \frac{50}{3} ms^{-1}r = 0.1km = 100m$ So $\tan \theta = \frac{\left(\frac{50}{3}\right)^2}{100\times 10} = \frac{5}{18}\quad \Rightarrow \quad \theta = \tan^{-1}\left(\frac{5}{18}\right)$
Q152
Allen
2. Dynamics of Circular Motion
Numerical
When an airplane is driven forward by its engines, the air flowing over its wings generates an upward force called lift. When this lift balances the weight of the plane, it can fly horizontally. A plane banks, that is, tilts its wings to the side, to turn, thus using the lift force another way. Note that it must then increase speed to generate more lift if it wants to stay at the same altitude while turning.
A jet airliner remains at constant altitude at a constant speed of v = 720 km/hr while banking at with respect to the horizon to turn in a circle. The radius (in km) of the circle in which the plane moves is given by R. Find the value of $\sqrt{R}$ .
Correct Answer: 4
Explanation:
Ans. 4
$F\cos \theta = mg$ ... (i)
and $F\sin \theta = \frac{mv^2}{R}$ ...(ii)
By solving equation (i) and (ii)
$R = 1 6 \mathrm{km}$
$F\cos \theta = mg$ ... (i)
and $F\sin \theta = \frac{mv^2}{R}$ ...(ii)
By solving equation (i) and (ii)
$R = 1 6 \mathrm{km}$
Q153
Allen
2. Dynamics of Circular Motion
Numerical
A mosquito is sitting on an L.P. record of a gramophone disc rotating on a turn table at $33\frac{1}{3}$ revolution per minute. The distance of the mosquito from the center of the disc is 10 cm. Find minimum friction coefficient between the record and the mosquito.
Take $g = 10 \, m/s^{2}$ .
Correct Answer: $\frac{\pi^2}{81}$
Explanation:
Ans. $\frac{\pi^2}{81}$
$n = 3 3 \frac {1}{3} r e v / m i n$
$\omega = \frac {2 \pi n}{6 0} = \frac {2 \pi \times 3 3 \frac {1}{3}}{6 0} \mathrm{rad/sec.,}$
r = 1 0 c m = 0.1 m.
For no slipping,
$f \geq m \omega^ {2} r$
$\mu m g \geq m \omega^ {2} r$
$\mu \geq \frac {\omega^ {2} r}{g}$
$\mu \geq \left(\frac {2 \pi 1 0 0}{6 0 \times 3}\right) ^ {2} \frac {0.1}{1 0}$
$\mu \geq \frac {\pi^ {2}}{8 1}$
$n = 3 3 \frac {1}{3} r e v / m i n$
$\omega = \frac {2 \pi n}{6 0} = \frac {2 \pi \times 3 3 \frac {1}{3}}{6 0} \mathrm{rad/sec.,}$
r = 1 0 c m = 0.1 m.
For no slipping,
$f \geq m \omega^ {2} r$
$\mu m g \geq m \omega^ {2} r$
$\mu \geq \frac {\omega^ {2} r}{g}$
$\mu \geq \left(\frac {2 \pi 1 0 0}{6 0 \times 3}\right) ^ {2} \frac {0.1}{1 0}$
$\mu \geq \frac {\pi^ {2}}{8 1}$
Q154
Allen
2. Dynamics of Circular Motion
Numerical
A motorcyclist wants to drive on the vertical surface of wooden ‘well’ of radius 5 m, in horizontal plane with speed of $5\sqrt{5}$ m/s. Find the minimum value of coefficient of friction between the tyres and the wall of the well. (take $g = 10 \, m/s^{2}$ )
Correct Answer: $\frac{2}{5}$
Explanation:
Ans. $\frac{2}{5}$
$N = \frac {m v ^ {2}}{r}$
Given $r = 5 \, m$ , $v = 5\sqrt{5} \, m/s$
for no slipping,
$f \geq m g$
$\mu_ {\mathrm{min}} N = m g$
$\mu_ {\mathrm{min}} = \frac {m g}{N} = \frac {r g}{v ^ {2}}$
$\mu_ {\mathrm{min}} = \frac {5 \times 1 0}{(5 \sqrt {5}) ^ {2}} = \frac {2}{5}$
$N = \frac {m v ^ {2}}{r}$
Given $r = 5 \, m$ , $v = 5\sqrt{5} \, m/s$
for no slipping,
$f \geq m g$
$\mu_ {\mathrm{min}} N = m g$
$\mu_ {\mathrm{min}} = \frac {m g}{N} = \frac {r g}{v ^ {2}}$
$\mu_ {\mathrm{min}} = \frac {5 \times 1 0}{(5 \sqrt {5}) ^ {2}} = \frac {2}{5}$
Q155
Allen
2. Dynamics of Circular Motion
Numerical
A train has to negotiate a curve of radius 400 m. By how much height should the outer rail be raised with respect to inner rail for a speed of 48 km/hr? The distance between the rails is 1 m :
Correct Answer: $\frac{2}{45} m$
Explanation:
Ans. $\frac{2}{45} m$ $v = 48 km/hr = 40/3 m/s$ For safe turn without friction $\tan \theta = \frac{v^2}{rg} = \frac{h}{x}$ Given $x \approx \ell = 1m$ $\Rightarrow h = \frac{v^2}{rg} = \frac{(40/3)^2}{400 \times 10} = \frac{2}{45} m$
Q156
Allen
2. Dynamics of Circular Motion
Numerical
A turn of radius 20 m is banked for the vehicles going at a speed of 36 km/h. If the coefficient of static friction between the road and the tyre is 0.4, what are the possible speeds of a vehicle so that it neither slips down nor skids up?
Between $\sqrt{\frac{50}{3}}\times\frac{18}{5}=14.7$ km/h and 54 km/hr
Correct Answer: $\sqrt{\frac{50}{3}}\times \frac{18}{5} = 14.7km / h$
Explanation:
Ans. $\sqrt{\frac{50}{3}} \times \frac{18}{5} = 14.7 \, \text{km/h}$
$v = 3 6 \mathrm{km} / h r = 1 0 \mathrm{m} / \mathrm{s}$
$\tan \theta = \frac {v ^ {2}}{R g} = \frac {1 0 \times 1 0}{2 0 \times 1 0} = \frac {1}{2}$
$v _ {\max} = \sqrt {R g \left(\frac {\tan \theta + \mu}{1 - \mu \tan \theta}\right)} = 1 5 m / s = 5 4 k m / h r \quad \text { Ans. }$
$\begin{array}{r l} v _ {\min} & = \sqrt {R g \left(\frac {\tan \theta - \mu}{1 + \mu \tan \theta}\right)} = \sqrt {\frac {2 0 \times 1 0 (0.5 - 0.4)}{(1 + 0.5 \times 0.4)}} \\ & = \sqrt {\frac {2 0 \times 1 0 \times 0.1}{1.2}} = \sqrt {\frac {5 0}{3}} \times \frac {1 8}{5} k m / h r \mathbf {A n s}. \end{array}$
$v = 3 6 \mathrm{km} / h r = 1 0 \mathrm{m} / \mathrm{s}$
$\tan \theta = \frac {v ^ {2}}{R g} = \frac {1 0 \times 1 0}{2 0 \times 1 0} = \frac {1}{2}$
$v _ {\max} = \sqrt {R g \left(\frac {\tan \theta + \mu}{1 - \mu \tan \theta}\right)} = 1 5 m / s = 5 4 k m / h r \quad \text { Ans. }$
$\begin{array}{r l} v _ {\min} & = \sqrt {R g \left(\frac {\tan \theta - \mu}{1 + \mu \tan \theta}\right)} = \sqrt {\frac {2 0 \times 1 0 (0.5 - 0.4)}{(1 + 0.5 \times 0.4)}} \\ & = \sqrt {\frac {2 0 \times 1 0 \times 0.1}{1.2}} = \sqrt {\frac {5 0}{3}} \times \frac {1 8}{5} k m / h r \mathbf {A n s}. \end{array}$
Q157
Allen
2. Dynamics of Circular Motion
Numerical
A circular racetrack of radius 300 m is banked at an angle of $15^{\circ}$ . If the coefficient of friction between the wheels of a race car and the road is 0.2.
(A) What is the optimum speed of the race car to avoid wear and tear on its tires?
(B) What is the maximum permissible speed to avoid slipping?
Correct Answer: (a) $28m / s$ (b) $38.19m / s$
Explanation:
Ans. (a) 28 m/s (b) 38.19 m/s
a) Optimum speed of car is $\tan\theta=\frac{v^{2}}{rg}$ $v^{2}=rg\tan\theta$ $=10\times300\times\tan15$ $v^{2}=780\ m/s$ v=28 m/s
b) Maximum permissible speed to avoid slipping is
$\begin{array}{r l} v _ {m} & = \sqrt {R g \left(\frac {\mu + \tan \theta}{1 - \mu \tan \theta}\right)} \\ & = \sqrt {3 0 0 \times 1 0 \left(\frac {0 . 2 + \tan 1 5}{1 - 0 . 2 \tan 1 5}\right)} \\ & = \sqrt {1 4 5 8 . 7} \\ & = 3 8. 1 9 m / s \end{array}$
a) Optimum speed of car is $\tan\theta=\frac{v^{2}}{rg}$ $v^{2}=rg\tan\theta$ $=10\times300\times\tan15$ $v^{2}=780\ m/s$ v=28 m/s
b) Maximum permissible speed to avoid slipping is
$\begin{array}{r l} v _ {m} & = \sqrt {R g \left(\frac {\mu + \tan \theta}{1 - \mu \tan \theta}\right)} \\ & = \sqrt {3 0 0 \times 1 0 \left(\frac {0 . 2 + \tan 1 5}{1 - 0 . 2 \tan 1 5}\right)} \\ & = \sqrt {1 4 5 8 . 7} \\ & = 3 8. 1 9 m / s \end{array}$
Q158
Allen
2. Dynamics of Circular Motion
Numerical
A block of mass m = 20 kg is kept at a distance R = 1m from central axis of rotation of a round turn table (A table whose surface can rotate about central axis). Table starts from rest and rotates with constant angular acceleration, $\alpha = 3$ rad/sec $^{2}$ . The friction coefficient between block and table is $\mu = 0.5$ . At time $t = \frac{x}{3}$ sec from starting of motion (i.e. t = 0 sec) the block is just about to slip. Find the value of x.
Correct Answer: 2
Explanation:
Ans. 2
$\mu m g = m \sqrt {a _ {t} ^ {2} + a _ {n} ^ {2}}$
$r \omega^ {2} = (\alpha t) ^ {2} = x ^ {2}$
$(0.5 \times 1 0) ^ {2} = x ^ {4} + 9$
$x ^ {4} = 1 6$
x = 2
$\mu m g = m \sqrt {a _ {t} ^ {2} + a _ {n} ^ {2}}$
$r \omega^ {2} = (\alpha t) ^ {2} = x ^ {2}$
$(0.5 \times 1 0) ^ {2} = x ^ {4} + 9$
$x ^ {4} = 1 6$
x = 2
Q159
Allen
2. Dynamics of Circular Motion
Numerical
A car goes on a horizontal circular road of radius R, the speed increasing at constant rate $\frac{dv}{dt}=a$ . The friction coefficient between the road and the tyre is $\mu$ . Find the speed at which the car will skid.
Correct Answer: $V = \left[\left(\mu^2 g^2 - a^2\right)R^2\right]^{\frac{1}{4}}$
Explanation:
Ans. $V = \left[\left(\mu^2 g^2 - a^2\right) R^2\right]^{\frac{1}{4}}$
$a _ {R} = \sqrt {a _ {N} ^ {2} + a _ {t} ^ {2}}$
$a _ {N} = \frac {V ^ {2}}{R}, a _ {t} = \frac {d V}{d t} = a$
Friction, $f = ma_{R}$
$\mu N = m \sqrt {\left(\frac {V ^ {2}}{R}\right) ^ {2} + \left(\frac {d V}{d t}\right) ^ {2}}$
$\mu N = m \sqrt {\frac {V ^ {4}}{R ^ {2}} + a ^ {2}}$
$V = \left[ \left(\mu^ {2} g ^ {2} - a ^ {2}\right) R ^ {2} \right] ^ {\frac {1}{4}}$
$a _ {R} = \sqrt {a _ {N} ^ {2} + a _ {t} ^ {2}}$
$a _ {N} = \frac {V ^ {2}}{R}, a _ {t} = \frac {d V}{d t} = a$
Friction, $f = ma_{R}$
$\mu N = m \sqrt {\left(\frac {V ^ {2}}{R}\right) ^ {2} + \left(\frac {d V}{d t}\right) ^ {2}}$
$\mu N = m \sqrt {\frac {V ^ {4}}{R ^ {2}} + a ^ {2}}$
$V = \left[ \left(\mu^ {2} g ^ {2} - a ^ {2}\right) R ^ {2} \right] ^ {\frac {1}{4}}$
Q160
Allen
2. Dynamics of Circular Motion
Numerical
A motorcycle has to move with a constant speed on an overbridge which is in the form of a circular are of radius R and has a total length L. Suppose the motorcycle starts from the highest point.
(A) What can its maximum velocity be for which the contact with the road is not broken at the highest point?
(B) If the motorcycle goes at speed $1/\sqrt{2}$ times the maximum found in part a), where will it lose the contact with the road?
(C) What maximum uniform speed can it maintain on the bridge if it does not lose contact anywhere on the bridge?
Correct Answer: (a) $v = \sqrt{Rg}$ , (b) $\frac{\pi R}{3}$ , (c) $V = \sqrt{gR\cos\left(\frac{L}{2R}\right)}$
Explanation:
Ans. a) $v = \sqrt{Rg}$ , b) $\frac{\pi R}{3}$ , c) $V = \sqrt{gR\cos\left(\frac{L}{2R}\right)}$
a) At highest point
$m g = \frac {m v ^ {2}}{R}$
$\Rightarrow v = \sqrt {R g}$
b) $V = \frac{1}{\sqrt{2}} \sqrt{Rg}$
Losses contact at angle $\theta$
$m g \cos \theta = \frac {m V ^ {2}}{R}$
c) $\alpha=\frac{L}{2R}$
So, $\frac{mV^2}{R} = mg\cos \theta$
$V = \sqrt {g R \cos \left(\frac {L}{2 R}\right)}$
a) At highest point
$m g = \frac {m v ^ {2}}{R}$
$\Rightarrow v = \sqrt {R g}$
b) $V = \frac{1}{\sqrt{2}} \sqrt{Rg}$
Losses contact at angle $\theta$
$m g \cos \theta = \frac {m V ^ {2}}{R}$
c) $\alpha=\frac{L}{2R}$
So, $\frac{mV^2}{R} = mg\cos \theta$
$V = \sqrt {g R \cos \left(\frac {L}{2 R}\right)}$
Q161
Allen
2. Dynamics of Circular Motion
Numerical
A block of mass m is kept on a horizontal ruler. The friction coefficient between the ruler and the block is $\mu$ . The ruler is fixed at one end and the block is at a distance L from the fixed end. The ruler is rotated about the fixed end in the horizontal plane through the fixed end. (a) What can the maximum angular speed be for which the block does not slip? (b) If the angular speed of the ruler is uniformly increased from zero at an angular acceleration $\alpha$ , at what angular speed will the block slip?
Correct Answer: (A) $\omega_{\max} = \sqrt{\frac{\mu g}{L}}$ , (B) $\omega = \left(\left(\frac{\mu g}{L}\right)^2 - \alpha^2\right)^{\frac{1}{4}}$
Explanation:
Ans. (A) $\omega_{\max} = \sqrt{\frac{\mu g}{L}}$ , B) $\omega = \left(\left(\frac{\mu g}{L}\right)^2 - \alpha^2\right)^{\frac{1}{4}}$
A)
$f _ {\max} = \mu N = (\mu m g)$
$f = (m \omega^ {2} L)$
$f \leq f _ {\mathrm{max}}$
$m \omega^ {2} L \leq \mu m g$
$\omega^ {2} \leq \frac {\mu g}{L} \Rightarrow \omega \leq \left(\sqrt {\frac {\mu g}{L}}\right)$
$\omega_ {\mathrm{max}} = \sqrt {\frac {\mu g}{L}}$
B)
$a _ {n e t} = \sqrt {a _ {C} ^ {2} + a _ {t} ^ {2}} = \sqrt {\left(\omega^ {2} L\right) ^ {2} + (\alpha L) ^ {2}}$
$f = m a _ {n e t} = m \sqrt {\left(\omega^ {2} L\right) ^ {2} + (\alpha L) ^ {2}}$
$f \leq f _ {\mathrm{max}}$
$f = f _ {\mathrm{max}}$
$m \sqrt {\left(\omega^ {2} L\right) ^ {2} + \left(\alpha L\right) ^ {2}} = \mu m g$
$\omega \left(\left(\frac {\mu g}{L}\right) ^ {2} - \alpha^ {2}\right) ^ {\frac {1}{4}}$
A)
$f _ {\max} = \mu N = (\mu m g)$
$f = (m \omega^ {2} L)$
$f \leq f _ {\mathrm{max}}$
$m \omega^ {2} L \leq \mu m g$
$\omega^ {2} \leq \frac {\mu g}{L} \Rightarrow \omega \leq \left(\sqrt {\frac {\mu g}{L}}\right)$
$\omega_ {\mathrm{max}} = \sqrt {\frac {\mu g}{L}}$
B)
$a _ {n e t} = \sqrt {a _ {C} ^ {2} + a _ {t} ^ {2}} = \sqrt {\left(\omega^ {2} L\right) ^ {2} + (\alpha L) ^ {2}}$
$f = m a _ {n e t} = m \sqrt {\left(\omega^ {2} L\right) ^ {2} + (\alpha L) ^ {2}}$
$f \leq f _ {\mathrm{max}}$
$f = f _ {\mathrm{max}}$
$m \sqrt {\left(\omega^ {2} L\right) ^ {2} + \left(\alpha L\right) ^ {2}} = \mu m g$
$\omega \left(\left(\frac {\mu g}{L}\right) ^ {2} - \alpha^ {2}\right) ^ {\frac {1}{4}}$
Q162
Allen
2. Dynamics of Circular Motion
Numerical
A track consists of two circular parts ABC and CDE of equal radius 100 m and joined smoothly as shown in figure. Each part subtends a rights angle at its centre. A cycle weighing 100 kg together with the rider travels at a constant speed of 18 km/h on the track.
(A) Find the normal contact force by the road on the cycle when it is at B and at D.
(B) Find the force of friction exerted by the track on the tyres when the cycle is at B, C, and D.
(C) Find the normal force between the cycle crosses C.
(D) What should be the minimum friction coefficient between the road and the tyre, which will ensure that the cyclist can move with constant speed? (Take $g = 10 \, m/s^2$ )
(A) Find the normal contact force by the road on the cycle when it is at B and at D.
(B) Find the force of friction exerted by the track on the tyres when the cycle is at B, C, and D.
(C) Find the normal force between the cycle crosses C.
(D) What should be the minimum friction coefficient between the road and the tyre, which will ensure that the cyclist can move with constant speed? (Take $g = 10 \, m/s^2$ )
Correct Answer: (a) 975 N, 1025 N (b) 707 N (c) 683 N (d) $\mu = 1.037$
Explanation:
Ans. (a) 975 N, 1025 N (b) 707 N (c) 683 N (d) $\mu = 1.037$
a) At $B, N = mg - \frac{mv^2}{R}$ At $D, N = mg + \frac{mv^2}{R}$
b) Friction force is zero (no sliding tendency)
At C, mg sin θ = F
c) Just before, $mg\cos \theta - N = \frac{mv^2}{R}$
d) Minimum coefficient of friction $\Rightarrow \mu N = mg\sin \theta$
a) At $B, N = mg - \frac{mv^2}{R}$ At $D, N = mg + \frac{mv^2}{R}$
b) Friction force is zero (no sliding tendency)
At C, mg sin θ = F
c) Just before, $mg\cos \theta - N = \frac{mv^2}{R}$
d) Minimum coefficient of friction $\Rightarrow \mu N = mg\sin \theta$
Q163
Allen
2. Dynamics of Circular Motion
Numerical
A table with smooth horizontal surface is turning at an angular speed $\omega$ about its axis. A groove is made on the surface along a radius and a particle is gently placed inside the groove at a distance a from the centre. Find the speed of the particle as its distance from the centre becomes L.
Correct Answer: $v = \omega \sqrt{L^2 - a^2}$
Explanation:
Ans. $v = \omega \sqrt{L^{2} - a^{2}}$
The situation is shown in figure.
Thus, the acceleration along the X-axis is
$a = \frac {F}{m} = \frac {m \omega^ {2} x}{m} = \omega^ {2} x$
or $\frac{dv}{dt} = \omega^2 x$
or $\frac{dv}{dx}\cdot \frac{dx}{dt} = \omega^2 x$
or $\frac{dv}{dx}\cdot v = \omega^2 x$
or $v dv = \omega^2 x dx$
or $\int_0^v v dv = \int_a^L\omega^2 x dx$
or $\left[\frac{1}{2}v^{2}\right]_{0}^{v}=\left[\frac{1}{2}\omega^{2}x^{2}\right]_{a}^{L}$
or $\frac{v^2}{2} = \frac{1}{2}\omega^2 (L^2 -a^2)$
or $v = \omega \sqrt{L^2 - a^2}$
The situation is shown in figure.
Thus, the acceleration along the X-axis is
$a = \frac {F}{m} = \frac {m \omega^ {2} x}{m} = \omega^ {2} x$
or $\frac{dv}{dt} = \omega^2 x$
or $\frac{dv}{dx}\cdot \frac{dx}{dt} = \omega^2 x$
or $\frac{dv}{dx}\cdot v = \omega^2 x$
or $v dv = \omega^2 x dx$
or $\int_0^v v dv = \int_a^L\omega^2 x dx$
or $\left[\frac{1}{2}v^{2}\right]_{0}^{v}=\left[\frac{1}{2}\omega^{2}x^{2}\right]_{a}^{L}$
or $\frac{v^2}{2} = \frac{1}{2}\omega^2 (L^2 -a^2)$
or $v = \omega \sqrt{L^2 - a^2}$
Q164
Allen
3. Work
Numerical
A sleeve of mass 2 kg at origin can move on wire of parabolic shape $x^{2} = 4y$ . Two forces $F_{1} = 6N$ and $F_{2} = 8N$ are applied on the sleeve. $F_{1}$ is constant and is in x-direction. $F_{2}$ is constant in direction and magnitude. Body is displaced from origin to x = 4, then net work done by $F_{1}$ and $F_{2}$ is
Correct Answer: 67.7 J
Explanation:
Ans. 67.7 J
$W _ {1} = \vec {F} _ {1}. \vec {d} _ {1} = (6) (4) = 2 4 N$
$W_{2} = \vec{F}_{2}.\vec{d}_{2}$ where $= 8\cos 30i + 8\sin 30j$ $= (4\sqrt{3}\hat{i} +4\hat{j}),\vec{s} = (4\hat{i} +4\hat{j})$
$\Rightarrow W_{2} = 43.7J$ therefore $W_{\mathrm{net}} = 67.7J$
$W _ {1} = \vec {F} _ {1}. \vec {d} _ {1} = (6) (4) = 2 4 N$
$W_{2} = \vec{F}_{2}.\vec{d}_{2}$ where $= 8\cos 30i + 8\sin 30j$ $= (4\sqrt{3}\hat{i} +4\hat{j}),\vec{s} = (4\hat{i} +4\hat{j})$
$\Rightarrow W_{2} = 43.7J$ therefore $W_{\mathrm{net}} = 67.7J$
Q165
Allen
3. Work
Numerical
A particle is subject to a force $F_{x}$ that varies with position as in figure. Find the work done by the force on the body as it moves (a) from x = 0 to x = 5.00 m, (b) from x = 5.00 m to x = 10.0 m, and (c) from x = 10.0 m to x = 15.0 m. (d) What is the total work done by the force over the distance x = 0 to x = 15.0 m?
Correct Answer: (a) 7.5 J; (b) 15 J; (c) 7.5 J; (d) 30 J
Explanation:
Ans. (a) 7.5 J; (b) 15 J; (c) 7.5 J; (d) 30 J
W = area under the graph.
W = area under the graph.
Q166
Allen
3. Work
Numerical
A spring, which is initially in its unstretched condition, is first stretched by a length x and then again by a further length x. The work done in the first case is $W_{1}$ and in the second case is $W_{2}$ . Find $\frac{W_{2}}{W_{1}}$ .
Correct Answer: 3
Explanation:
Ans. 3
$W _ {1} = \frac {1}{2} k x ^ {2}$
$W_{2} = \frac{1}{2} k\left[4x^{2} - x^{2}\right] = \frac{3}{2} kx^{2}$
$W _ {1} = \frac {1}{2} k x ^ {2}$
$W_{2} = \frac{1}{2} k\left[4x^{2} - x^{2}\right] = \frac{3}{2} kx^{2}$
Q167
Allen
5. Work Energy Theorem
Numerical
Position-time graph of a particle of mass 2 kg is shown in figure. Total work done on the particle from t = 0 to t = 4s is
Correct Answer: 0
Explanation:
Ans. 0
Total work done = change in kinetic energy
$= \frac {1}{2} m v _ {2} ^ {2} - \frac {1}{2} m v _ {1} ^ {2} = \frac {1}{2} (2) \times (1) ^ {2} - \frac {1}{2} \times 2 \times (1) ^ {2} = 0$
Total work done = change in kinetic energy
$= \frac {1}{2} m v _ {2} ^ {2} - \frac {1}{2} m v _ {1} ^ {2} = \frac {1}{2} (2) \times (1) ^ {2} - \frac {1}{2} \times 2 \times (1) ^ {2} = 0$
Q168
Allen
5. Work Energy Theorem
Numerical
A mass m slides from rest at height h down a smooth curved surface which becomes horizontal at zero height (see figure). A spring is fixed horizontally on the level part of the surface. The spring constant is k N/m. When the mass encounters the spring it compresses it by an amount x = h/10. If m = 1 kg, h = 5m then find k /100.
Correct Answer: 4
Explanation:
Ans. 4
$m g h = \frac {1}{2} k x ^ {2}$
$1 \times 10 \times 5 = \frac{1}{2} \times k \times \frac{1}{4}$
$m g h = \frac {1}{2} k x ^ {2}$
$1 \times 10 \times 5 = \frac{1}{2} \times k \times \frac{1}{4}$
Q169
Allen
5. Work Energy Theorem
Numerical
A small block of mass m is lying at rest at point P of a wedge having a smooth semi-circular track of radius R. The minimum value of horizontal acceleration $a_{0}$ (in $m/s^{2}$ ) of wedge so that mass can just reach the point Q, is
Correct Answer: 10
Explanation:
Ans. 10
By applying work energy theorem in wedge frame between point P and Q. $\Delta KE = W_g + W_{pseudo force}$ $\Rightarrow$ 0 = -mgR + maR $\Rightarrow$ a = g
By applying work energy theorem in wedge frame between point P and Q. $\Delta KE = W_g + W_{pseudo force}$ $\Rightarrow$ 0 = -mgR + maR $\Rightarrow$ a = g
Q170
Allen
6. Potential Energy
Numerical
A uniform chain of mass m and length l overhangs a table with its two third part on the table. Find the work to be done by a person to put the hanging part back on the table.
Correct Answer: $\frac{mgl}{18}$
Explanation:
Ans. $\frac{mgl}{18}$ Mass of dx = dx × $\frac{m}{L}$ dW = dmgh $dW = dx \times \frac{m}{L}gx$ $\int_{0}^{w} dW = \int_{0}^{L/3} \frac{m}{L}gxdx = \frac{mgL}{18}$
Q171
Allen
6. Potential Energy
Numerical
A uniform chain of length L and mass M overhangs a horizontal table with its two third part on the table. The friction coefficient between the table and the chain is $\mu$ . Find the work done by the friction during the period the chain slips off the table.
Correct Answer: $-\frac{2}{9}\mu MgL$
Explanation:
Ans. $-\frac{2}{9}\mu MgL$ $dM = \frac{M}{L}dx$ $\int_{0}^{w} dW = \int_{0}^{2L/3} -\mu \frac{M}{L}dxgx = \left( \frac{-2\mu MgL}{9} \right)$
Q172
Allen
8. Power
Numerical
A 4 kg particle moves along the X-axis. It's position x varies with time according to $x(t) = t + 2t^{3}$ , where x is in m and t is in seconds. Compute:
(I) The kinetic energy at time t.
(II) The force acting on the particle at time t.
(III) The power delivered to the particle at time t.
(IV) The work done on the particle from t = 0 to t = 2 seconds.
(I) The kinetic energy at time t.
(II) The force acting on the particle at time t.
(III) The power delivered to the particle at time t.
(IV) The work done on the particle from t = 0 to t = 2 seconds.
Correct Answer: (i) $2 + 24\mathrm{t}^2 +72\mathrm{t}^4 J,$ (ii) $48tN,$ (iii) $48\mathrm{t} + 288\mathrm{t}^3 W,$ (iv) $1248J$
Explanation:
Ans. (i) $2 + 24t^2 + 72t^4J$ , (ii) 48 t N, (iii) 48t + 288t³W, (iv) 1248J
v = 1 + 6t²
and a = 12t $KE = \frac{1}{2}m\left(1 + 36t^4 + 12t^2\right)$ F = 4(12 t) = 48t
P = F·v
v = 1 + 6t²
and a = 12t $KE = \frac{1}{2}m\left(1 + 36t^4 + 12t^2\right)$ F = 4(12 t) = 48t
P = F·v
Q173
Allen
8. Power
Numerical
The elevator E has a mass of 3000 kg when fully loaded and is connected as shown to a counterweight W of mass 1000 kg. Determine the power in kilowatts delivered by the motor:
(A) when the elevator is moving down at a constant speed of 3 m/s,
(B) when it has an upward velocity of 3 m/s and a deceleration of $0.5 \, m/s^{2}$ .
(A) when the elevator is moving down at a constant speed of 3 m/s,
(B) when it has an upward velocity of 3 m/s and a deceleration of $0.5 \, m/s^{2}$ .
Correct Answer: (a) $-30\mathrm{kW}$ , (b) $19.5\mathrm{Kw}$
Explanation:
Ans. (a) -30 kW, (b) 19.5 kW
(a) P = F·v = -T × 6
= -5000 × 6
= -30 kW
(a) P = F·v = -T × 6
= -5000 × 6
= -30 kW
Q174
Allen
8. Power
Numerical
Power applied to a particle varies with time as $P = (3t^{2} - 2t + 1)$ watt, where t is in second. Find the change in its kinetic energy between time t = 2 s and t = 4 s.
Correct Answer: $46J$
Explanation:
Ans. 46J
$\begin{array}{r l} & P = F \cdot v = m v \cdot \frac {d v}{d t} \\ & \Rightarrow \quad \frac {1}{2} m (v ^ {2} - u ^ {2}) = \int P d t \\ & \qquad = [ t ^ {3} - t ^ {2} + t ] _ {2} ^ {4} \end{array}$
$\begin{array}{r l} & P = F \cdot v = m v \cdot \frac {d v}{d t} \\ & \Rightarrow \quad \frac {1}{2} m (v ^ {2} - u ^ {2}) = \int P d t \\ & \qquad = [ t ^ {3} - t ^ {2} + t ] _ {2} ^ {4} \end{array}$
Q175
Allen
9. MISCELLANEOUS QUESTION
Numerical
A block of mass m moves on a horizontal circle against the wall of a cylindrical room of radius R. The floor of the room on which the block moves is smooth but the friction coefficient between the wall and the block is $\mu$ . The block is given and initial speed $v_{0}$ . As a function of the speed v write
(A) the normal force by the wall on the block, (b) the frictional force by the wall and (c) the tangential acceleration of the block. (d) Integrate the tangential acceleration $\left(\frac{dv}{dt}=v\frac{dv}{ds}\right)$ to obtain the speed of the block after one revolution.
(A) the normal force by the wall on the block, (b) the frictional force by the wall and (c) the tangential acceleration of the block. (d) Integrate the tangential acceleration $\left(\frac{dv}{dt}=v\frac{dv}{ds}\right)$ to obtain the speed of the block after one revolution.
Correct Answer: (a) $\frac{mv^2}{R}$ ; (b) $\frac{\mu mv^2}{R}$ ; (c) $-\frac{\mu v^2}{R}$ ; (d) $v_0 e^{-2\pi \mu}$
Explanation:
Ans. (a) $\frac{mv^{2}}{R}$ ; (b) $\frac{\mu mv^{2}}{R}$ ; (c) $-\frac{\mu v^{2}}{R}$ ; (d) $v_{0}e^{-2\pi\mu}$
Top vies
a) $F_{c}=N$ $F_{t}=f$ $N=\frac{mv^{2}}{R}$ and $f=\mu N$
b) $f=\frac{\mu mv^{2}}{R}$
c) $a_{t}=\frac{f_{k}}{m}=\left(\frac{\mu mv^{2}}{R}\right).\frac{1}{m}$ $\vec{a}_{t}=-\frac{\mu v^{2}}{R}$ (acc. is o
d) $v\frac{dv}{ds}=-\frac{\mu v^{2}}{R}$ $v\frac{dv}{ds}=-\frac{\mu v^{2}}{R}$ $\frac{dv}{ds}=-\frac{\mu v}{R}$ $\int_{0}^{2\pi R}-\frac{\mu}{R}ds=\int_{v_{0}}^{v}\frac{1}{v}dv$ $-\frac{\mu}{R}(2\pi R)=\ln\left(\frac{v}{v_{0}}\right)$ $v=v_{0}e^{-2\pi\mu}$
$\vec{a}_t = -\frac{\mu v^2}{R}$ (acc. is opposite to velocity)
Top vies
a) $F_{c}=N$ $F_{t}=f$ $N=\frac{mv^{2}}{R}$ and $f=\mu N$
b) $f=\frac{\mu mv^{2}}{R}$
c) $a_{t}=\frac{f_{k}}{m}=\left(\frac{\mu mv^{2}}{R}\right).\frac{1}{m}$ $\vec{a}_{t}=-\frac{\mu v^{2}}{R}$ (acc. is o
d) $v\frac{dv}{ds}=-\frac{\mu v^{2}}{R}$ $v\frac{dv}{ds}=-\frac{\mu v^{2}}{R}$ $\frac{dv}{ds}=-\frac{\mu v}{R}$ $\int_{0}^{2\pi R}-\frac{\mu}{R}ds=\int_{v_{0}}^{v}\frac{1}{v}dv$ $-\frac{\mu}{R}(2\pi R)=\ln\left(\frac{v}{v_{0}}\right)$ $v=v_{0}e^{-2\pi\mu}$
$\vec{a}_t = -\frac{\mu v^2}{R}$ (acc. is opposite to velocity)