Work, Power & Energy

A → (r)
B → (p)
C → (q, s)
D → (q, s)
Conceptual

A → (p, s)
B → (q, r)
C → (p, s)
D → (q, r)
Conceptual

$\begin{array}{r l}&{\mathrm{A} \rightarrow (\mathrm{s})}\\&{\mathrm{B} \rightarrow (\mathrm{r})}\\&{\mathrm{C} \rightarrow (\mathrm{p})}\\&{\mathrm{D} \rightarrow (\mathrm{q})}\\&{\frac {v ^ {2}}{R} = 9 t ^ {2}}\\&{\Rightarrow v ^ {2} = (1 6) \big (9 t ^ {2} \big)}\\&{\Rightarrow v = 1 2 t}\\&{\Rightarrow a _ {t} = \frac {d v}{d t} = 1 2}\\&{\Rightarrow F _ {t} = m a _ {t} = \frac {1}{2} (1 2) = 6 \mathrm{N}}\\&{\mathrm{At} t = 1 \mathrm{s}, a _ {n} = \frac {v ^ {2}}{R} = 9}\\&{\Rightarrow F _ {n} = m a _ {n} = \frac {9}{2} \mathrm{N}}\\&{\Rightarrow F = \sqrt {F _ {t} ^ {2} + F _ {n} ^ {2}} = \sqrt {3 6 + \frac {8 1}{4}}}\end{array}$
$\begin{array}{r l} \Rightarrow & F = \sqrt {\frac {2 2 5}{4}} = \frac {1 5}{2} = 7. 5 \mathrm{N} \\ & P = \vec {F} \cdot \vec {v} \\ \Rightarrow & P = (F _ {t}) (v) \\ \Rightarrow & P = (6) (1 2) = 7 2 \mathrm{watt} \\ & P _ {\mathrm{av}} = \frac {W}{t} \\ \Rightarrow & P _ {\mathrm{av}} = F _ {t} (v _ {\mathrm{av}}) \end{array}$
But $v_{av}=\frac{\int\limits_{0}^{1}vdt}{\int\limits_{0}^{1}dt}=\frac{12\int\limits_{0}^{1}tdt}{1}$ $\Rightarrow \quad v_{av}=\frac{12t^{2}}{2}\bigg|_{0}^{1}=6\ ms^{-1}$ $\Rightarrow \quad P_{av}=(6)(6)=36\ watt$

A → (s)
B → (r)
C → (q)
D → (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
$a = \frac {F}{m} = 5 \sqrt {3} \mathrm{ms} ^ {- 2}$

Since the block does not slide on the wedge. So
$f = m g \sin (4 5 ^ {\circ}) = \frac {(1) (1 0)}{\sqrt {2}} = \frac {1 0}{\sqrt {2}} \mathrm{N}$


$\begin{array}{r l} & {\mathrm{Now} W _ {f} = \vec {f} \cdot \vec {s}} \\ & {\Rightarrow W _ {f} = (f \cos 4 5 ^ {\circ}) (2)} \\ & {\Rightarrow W _ {f} = \bigg (\frac {1 0}{\sqrt {2}} \bigg) \bigg (\frac {1}{\sqrt {2}} \bigg) (2)} \\ & {\Rightarrow W _ {f} = 1 0 \mathrm{J}} \end{array}$

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.

(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, $v_{f} = 5 \, ms^{-1} = 500 \, cms^{-1}$

(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
$\begin{array}{r l} & m g (h) = \frac {1}{2} m (2 R g) + m g (2 R) \\ \Rightarrow & h = 3 R \\ \Rightarrow & * = 3 \end{array}$

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
$\Rightarrow H = (2. 5) (4 0) = 1 0 0 \mathrm{cm}$

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 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
$\begin{array}{r l} & (U + K) _ {\mathrm{atA}} = (U + K) _ {\mathrm{atB}} \\ \Rightarrow & \frac {1}{2} m u ^ {2} + m g h = \frac {1}{2} m v ^ {2} + 0 \\ \Rightarrow & v = \sqrt {u ^ {2} + 2 g h} = \sqrt {\frac {k x ^ {2}}{m} + 2 g h} \\ \text {Now,} & x = 5 \mathrm{cm} = 0.0 5 \mathrm{m} \\ & k = 2 4 0 0 \mathrm{Nm} ^ {- 1} \\ & m = 1 2 0 \mathrm{g} = 0.1 2 \mathrm{kg} \\ & h = 2. 5 \mathrm{m} \\ \Rightarrow & v = \sqrt {\frac {(2 4 0 0) (0.0 5) ^ {2}}{0.1 2} + 2 (1 0) (2.5)} \\ \Rightarrow & v = \sqrt {5 0 + 5 0} = 1 0 \mathrm{ms} ^ {- 1} \end{array}$

$(a) \text { Increase in thermal energy of block - floor system is }$
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

Let $\ell_{0}$ be the unstretched length of the spring, then by Law of Conservation of Energy, we have
$\begin{array}{r l} & (U + K) _ {\mathrm{atA}} = (U + K) _ {\mathrm{atB}} \\ & \frac {1}{2} k (3 - \ell_ {0}) ^ {2} + \frac {1}{2} m v _ {A} ^ {2} = \frac {1}{2} k (5 - \ell_ {0}) ^ {2} + \frac {1}{2} m v _ {B} ^ {2} \\ \text {But} v _ {B} = 0 \\ \Rightarrow & (3 - \ell_ {0}) ^ {2} + 4 = (5 - \ell_ {0}) ^ {2} \\ \Rightarrow & (9 + \not \ell_ {0} ^ {2} - 6 \ell_ {0}) + 4 = (2 5 + \not \ell_ {0} ^ {2} - 1 0 \ell_ {0}) \\ \Rightarrow & 1 6 - 4 \ell_ {0} = 4 \\ \Rightarrow & \ell_ {0} = 3 \mathrm{m} \end{array}$

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
$\begin{array}{r l} & {- \bigg (\frac {1}{5} \bigg) (3 0 0) \bigg (\frac {4}{5} \bigg) = - (3 0 0) \bigg (\frac {3}{5} \bigg) + (5 0) (2) +} \\ & {\qquad \qquad \qquad \qquad \qquad \frac {1}{2} (3 0) v ^ {2} + \frac {1}{2} (5) (2 v) ^ {2}} \\ {\Rightarrow} & {- 4 8 = - 1 8 0 + 1 0 0 + 2 5 v ^ {2}} \\ {\Rightarrow} & {v = \sqrt {\frac {3 2}{2 5}} = \frac {4 \sqrt {2}}{5}} \\ {\Rightarrow} & {* = 2} \end{array}$

$h = \ell \sin \theta = 2 \sin \theta$
$\begin{array}{r l} & v ^ {2} = v _ {0} ^ {2} + 2 g h = 2 5 + 2 \times 1 0 \times 2 \sin \theta \\ & \Rightarrow v ^ {2} = 2 5 + 4 0 \sin \theta \\ & \text {Further,} T - m g \sin \theta = \frac {m v ^ {2}}{\ell} \\ & \text {Given,} T = 2 m g \\ & \Rightarrow 2 m g - m g \sin \theta = \frac {m v ^ {2}}{\ell} \\ & \Rightarrow 2 (1 0) - 1 0 \sin \theta = \frac {2 5 + 4 0 \sin \theta}{2} \\ & \Rightarrow \sin \theta = \frac {1}{4} \\ & \Rightarrow \theta = \sin^ {- 1} \left(\frac {1}{4}\right) \\ & \text {So,} * = 4 \end{array}$

$\sin \theta = \frac{3}{5}$
At Q, T = 0 and hence the particle will become a projectile following a parabolic path.

$\begin{array}{r l} & {\Rightarrow \frac {m v ^ {2}}{r} = m g \sin \theta} \\ & {\Rightarrow v = \sqrt {r g \sin \theta} = \sqrt {(5) (1 0) \left(\frac {3}{5}\right)} = \sqrt {3 0} \mathrm{ms} ^ {- 1}} \\ & {\Rightarrow * = 3 0} \end{array}$
The maximum height of the projectile is
$\begin{array}{r l} & H = \frac {v ^ {2} \sin^ {2} (9 0 ^ {\circ} - \theta)}{2 g} = \frac {v ^ {2} \cos^ {2} \theta}{2 g} \\ \Rightarrow & H = \frac {(3 0) \left(\frac {1 6}{2 5}\right)}{2 (1 0)} = \frac {2 4}{2 5} \\ \Rightarrow & H = 0. 9 6 \mathrm{m} = 9 6 \mathrm{cm} \\ \Rightarrow & H = 2 4 x = 9 6 \\ \Rightarrow & x = 4 \end{array}$

Speed of each particle at angle $\theta$ using Law of Conservation of Energy is
$\begin{array}{r l} & v = \sqrt {2 g h}, \text {where} h = R (1 - \cos \theta) \\ \Rightarrow & v = \sqrt {2 g R (1 - \cos \theta)} \end{array}$
Also, at angle $\theta$ , we have
$\begin{array}{r l} & N + m g \cos \theta = \frac {m v ^ {2}}{R} \\ \Rightarrow & N + m g \cos \theta = 2 m g (1 - \cos \theta) \\ \Rightarrow & N = 2 m g - 3 m g \cos \theta \end{array}.....(i)$
When the tube just breaks its contact with the ground then
$2 N \cos \theta = M g$
Substituting the value of N from (1), we get
$4 m g \cos \theta - 6 m g \cos^ {2} \theta = M g$
Substituting, $\theta = 60^{\circ}$ , we get
$\begin{array}{r l} & 2 m g - \frac {3 m g}{2} = M g \\ \Rightarrow & \frac {m}{M} = 2 \end{array}$

Initially, the normal reaction on each ball will be radially outwards and later it will be radially inwards, so that the net normal reaction on the tube is radially outwards to break it off from the ground.
Let v be the speed of ball at point B. Then by Law of Conservation of Energy applied between A and B, we get
$U _ {A} + K _ {A} = U _ {B} + K _ {B}$
$\Rightarrow 0 + \frac {1}{2} (1) (5) ^ {2} = (1) (1 0) (1) + \frac {1}{2} (1) v ^ {2}$
$\Rightarrow \frac {v ^ {2}}{2} = \frac {2 5}{2} - 1 0\Rightarrow v ^ {2} = 5$
Force acting on particle at $B$ is
$F = \sqrt {\left(m g\right) ^ {2} + \left(\frac {m v ^ {2}}{r}\right) ^ {2}}$
$\Rightarrow F = \sqrt {(1 0) ^ {2} + \left(\frac {1 \times 5}{1}\right) ^ {2}}$
$\Rightarrow F = \sqrt {1 2 5}$
$\Rightarrow F = 5 \sqrt {5} \mathrm{N}$
$\Rightarrow x = 5$