Work, Power & Energy

Reference level at the centre of the sphere means gravitational potential energy is zero at centre.
Consider an element of length dx subtending an angle $d\theta$ at an angle $\theta$ with the vertical. The vertical height of this element is $R\cos\theta$ .

If dU be the potential energy of the element, then $dU = (dm)gh$ , where dm is mass of element.
$\begin{array}{l l} \Rightarrow & d m = \left(\frac {m}{\ell}\right) d x = \frac {m}{\ell} (R d \theta) \qquad \{\because d x = R d \theta \} \\ \text {So,} & d U = \frac {m}{\ell} (R d \theta) g (R \cos \theta) = \frac {m}{\ell} R ^ {2} g \cos \theta d \theta \\ \Rightarrow & U = \int d U = \frac {m g R ^ {2}}{\ell} \int_ {0} ^ {\theta_ {0}} \cos \theta d \theta = \frac {m g R ^ {2}}{\ell} \Big [ \sin \theta | _ {0} ^ {\theta_ {0}} \Big ] \\ \Rightarrow & U = \frac {m g R ^ {2}}{\ell} \big [ \sin \theta_ {0} \big ] = \frac {m g R ^ {2}}{\ell} \sin \left(\frac {\ell}{R}\right) \quad \Bigg \{\because \theta_ {0} = \frac {\ell}{R} \Bigg \} \end{array}$
Hence, the correct answer is (A).

$h = 2\ell \sin \alpha$
A is the lowest point and B the highest point. At B, in critical case tension is zero. Let velocity of particle at B at this instant be $v_{B}$ . Then
$m g \sin \alpha = \frac {m v _ {B} ^ {2}}{\ell}$
$\Rightarrow v_{B}^{2}=g\ell\sin\alpha$

Now by Law of Conservation of Energy, we have
$\begin{array}{r l} & v _ {A} ^ {2} = v _ {B} ^ {2} + 2 g h \\ \Rightarrow & v _ {A} ^ {2} = (g \ell \sin \alpha) + 2 g (2 \ell \sin \alpha) \\ \Rightarrow & v _ {A} = \sqrt {5 g \ell \sin \alpha} \end{array}$
Hence, the correct answer is (D).

Elongation in spring is
$x = \frac {1 3 l _ {0}}{1 2} - l _ {0} = \frac {l _ {0}}{1 2}$
Work done by lifting force is
$\begin{array}{r l} & W _ {\mathrm{ext}} = \Delta U + \Delta K = \Delta U _ {g} + \Delta U _ {e} + 0 \\ \Rightarrow & W _ {\mathrm{ext}} = \Delta U _ {g} + \Delta U _ {e} \\ \Rightarrow & W _ {\mathrm{ext}} = m g \left(\frac {5 l _ {0}}{1 2}\right) + \frac {1}{2} k \left(\frac {l _ {0}}{1 2}\right) ^ {2} \\ \Rightarrow & W = \frac {5 m g l _ {0}}{1 2} + \frac {k l _ {0} ^ {2}}{2 8 8} \end{array}$
Hence, the correct answer is (A).

If $R$ be the radius of the circumference, then
$\begin{array}{l l} & O P = O Q \cos 6 0 ^ {\circ} = (2 R) \left(\frac {1}{2}\right) = R \\ \Rightarrow & h _ {1} = O P \cos 6 0 ^ {\circ} = \frac {R}{2} \text { and } h _ {2} = 2 R \\ \Rightarrow & \frac {v _ {1}}{v _ {2}} = \frac {\sqrt {2 g h _ {1}}}{\sqrt {2 g h _ {2}}} = \sqrt {\frac {h _ {1}}{h _ {2}}} = \sqrt {\frac {1}{4}} = \frac {1}{2} \end{array}$
Hence, the correct answer is (C).

By Law of Conservation of Energy applied between A and P, we get
$\begin{array}{r l} & {\frac {1}{2} m v ^ {2} = m g R \sin \theta} \\ {\Rightarrow} & {v = \sqrt {2 g R \sin \theta}} \end{array}$
$\mathrm{Atpoint} P, N = m g \sin \theta + \frac {m v ^ {2}}{R}$
$\Rightarrow N = mg\sin \theta + 2mg\sin \theta = 3mg\sin \theta$
Centripetal force is given by
$F _ {C} = \frac {m v ^ {2}}{R} = 2 m g \sin \theta$
So, the required ratio is
$\frac {F _ {C}}{N} = \frac {2}{3}$
Hence, the correct answer is (C).

$\begin{array}{r l} & a = \left(\frac {m _ {2} - m _ {1} \sin \theta - \mu m _ {1} \cos \theta}{m _ {1} + m _ {2}}\right) g \\ & \Rightarrow \quad a = \left\{\frac {1 0 - 4 \sin 3 0 - (0 . 2) 4 \cos 3 0}{1 4} \right\} 1 0 \\ & \Rightarrow \quad a = 5. 2 \mathrm{ms} ^ {- 2} \\ & \text {Since,} v ^ {2} - 0 ^ {2} = 2 a s \\ & \Rightarrow \quad v ^ {2} = 2 (5. 2) (4) \\ & \Rightarrow \quad v \approx 6. 5 \mathrm{ms} ^ {- 1} \end{array}$
Hence, the correct answer is (B).

$m g \cos \theta - N = \frac {m v ^ {2}}{a}$
The body breaks off the sphere when N = 0
$\Rightarrow m g \cos \theta = \frac {m v ^ {2}}{a}$
$\Rightarrow v ^ {2} = g a \cos \theta\tag{... (1}$
Applying Law of Conservation of Mechanical Energy, we get
$\begin{array}{r l} & {\frac {1}{2} m v ^ {2} = m g h} \\ {\Rightarrow} & {v ^ {2} = 2 g h} \end{array}\tag{... (2}$
From (1) and (2), we get
$g a \cos \theta = 2 g h$
$\Rightarrow \cos \theta = 2 h / a$
Hence, the correct answer is (B).

$\begin{array}{l l} 5. & W = \int_ {(0, 0)} ^ {(1, 1)} \vec {F} \cdot d \vec {\ell} \\ & \text { where } d \vec {\ell} = d x \hat {i} + d y \hat {j} + d z \hat {k} \\ & \Rightarrow W = \int_ {(0, 0)} ^ {(1, 1)} (x ^ {2} d y + y d x) \\ & \Rightarrow W = \int_ {(0, 0)} ^ {(1, 1)} (y ^ {2} d y + x d x) \\ & \Rightarrow W = \left(\frac {y ^ {3}}{3} + \frac {x ^ {2}}{2}\right) \Big | _ {(0, 0)} ^ {(1, 1)} = \frac {5}{6} J \end{array} \quad \{\because x = y \}$
Hence, the correct answer is (B).

For the particle just to complete the vertical circle, speed at topmost point is
$v _ {0} = \sqrt {g r}$
Speed when the string becomes horizontal is
$\begin{array}{r l} & v = \sqrt {u ^ {2} + 2 g r} = \sqrt {g r + 2 g r} = \sqrt {3 g r} \\ \Rightarrow & v = \sqrt {3} v _ {0} \end{array}$
Tension, when string becomes horizontal, is
$\begin{array}{r l} & T = \frac {m v ^ {2}}{r} = \frac {m (\sqrt {3} v _ {0}) ^ {2}}{r} \\ \Rightarrow & T = \frac {3 m v _ {0} ^ {2}}{r} = \frac {3 m (g r)}{r} = 3 m g \end{array}$
Hence, the correct answer is (D).

From Work Energy Theorem, $W = \Delta K$
$\begin{array}{r l} \Rightarrow & P t = \frac {1}{2} m v ^ {2} \\ \Rightarrow & v = \sqrt {\frac {2 P t}{m}} \\ \Rightarrow & v = \frac {d s}{d t} = \sqrt {\frac {2 P}{m}} t ^ {1 / 2} \\ \Rightarrow & \int_ {0} ^ {s} d s = \sqrt {\frac {2 P}{m}} \int_ {0} ^ {t} t ^ {1 / 2} d t \\ \Rightarrow & s = \frac {2}{3} \sqrt {\frac {2 P}{m}} t ^ {3 / 2} \end{array}\tag{... (1}$
From equations (1) and (2), we get
$\begin{array}{r l} & {\frac {s}{v} = \frac {2}{3} t} \\ {\Rightarrow} & {\frac {s}{v} \propto t} \end{array}\tag{... (2}$
i.e., graph between $\frac{s}{v}$ and t is a straight line passing through origin.
Hence, the correct answer is (B).

$\begin{array}{r l} & u ^ {2} = g L (2 + 3 \sin \phi) = \frac {7}{2} g L \\ & \Rightarrow \quad \sin \phi = \frac {1}{2} \\ & \Rightarrow \quad \theta = 9 0 ^ {\circ} + \phi = 1 2 0 ^ {\circ} \end{array}$
Hence, the correct answer is (C).

The centre of mass of the rope is at a height $\frac{h}{2}$ . So, total work done is the sum of work done to pull the bucket of mass M to a height h and the work done to pull the CM of the rope of mass m to a height $\frac{h}{2}$ . So,
$W = M g h + m g \frac {h}{2}$
$\Rightarrow W = \left(M + \frac {m}{2}\right) g h$
Hence, the correct answer is (B).

Initial velocity = final velocity = 0
But displacement $\Delta x \neq 0$ {∵ $\Delta x = \text{Area under } v-t \text{ graph}$ }
From Work Energy Theorem, $W = \Delta KE = 0$ Hence, the correct answer is (B).

$a = \frac {| \vec {F} |}{m} = \frac {5}{2} = 2. 5 \mathrm{ms} ^ {- 2} \\ \text {Velocity at t = 20 s is v = at} \\ \Rightarrow \quad v = (2. 5) (2 0) = 5 0 \mathrm{ms} ^ {- 1} \\ \Rightarrow \quad \mathrm{K.E.} = \frac {1}{2} (2) (2 5 0 0) \\ \Rightarrow \quad \mathrm{K.E.} = 2 5 0 0 \mathrm{J} \\ \text {Hence, the correct answer is (C).} $

From Work Energy Theorem
$\begin{array}{r l} & {\Delta K E = W _ {\mathrm{net}}} \\ {\Rightarrow} & {K _ {f} - K _ {i} = \int P d t} \\ {\Rightarrow} & {\frac {1}{2} m v ^ {2} = \int_ {0} ^ {2} \biggl (\frac {3}{2} t ^ {2} \biggr) d t} \\ {\Rightarrow} & {v ^ {2} = \biggl (\frac {t ^ {3}}{2} \biggr) \bigg | _ {0} ^ {2}} \\ {\Rightarrow} & {v = 2 \mathrm{ms} ^ {- 1}} \end{array}$
$\left\{\because m = 2 \mathrm{kg} \right\}$
Hence, the correct answer is (B).

Work done by gravitational force is equal to the change in kinetic energy.
Let v be the velocity of particle at point B is v

The horizontal component of velocity remains constant, so
$u \cos \alpha = v \cos \left(\frac {\alpha}{2}\right)$
$\Rightarrow v = \frac {u \cos \alpha}{\cos (\alpha / 2)}$
$\begin{array}{l l} \Rightarrow & W = \frac {1}{2} m v ^ {2} - \frac {1}{2} m (u \cos \alpha) ^ {2} \\ \Rightarrow & W = \frac {1}{2} m \left[ \frac {u ^ {2} \cos^ {2} \alpha}{\cos^ {2} (\alpha / 2)} - u ^ {2} \cos^ {2} \alpha \right] \\ \Rightarrow & W = \frac {1}{2} m u ^ {2} \cos^ {2} \alpha \left[ \sec^ {2} \left(\frac {\alpha}{2}\right) - 1 \right] \\ \text { Since } & 1 + \tan^ {2} \theta = \sec^ {2} \theta \\ \Rightarrow & W = \frac {1}{2} m u ^ {2} \cos^ {2} \alpha \tan^ {2} \left(\frac {\alpha}{2}\right) \\ & \text { Hence, the correct answer is (C). } \end{array}$

Increase in P.E. is the increase in P.E. of C.M.
$\begin{array}{l l} & \Delta U = m g \ell_ {C M} (1 - \cos \theta) \\ \Rightarrow & \Delta U = \bigg (\frac {1 0 0}{1 0 0 0} \bigg) (1 0) \bigg (\frac {1}{2} \bigg) \bigg (1 - \frac {1}{2} \bigg) \\ \Rightarrow & \Delta U = 0. 2 5 \mathrm{J} \\ & \text {Hence, the correct answer is (A).} \end{array}$

$(K. E.) _ {\mathrm{man}} = \frac {1}{2} (\mathrm{K.E.}) _ {\mathrm{boy}} \\ \Rightarrow \quad \frac {1}{2} M v _ {m} ^ {2} = \frac {1}{2} \left[ \frac {1}{2} \left(\frac {M}{2}\right) v _ {b} ^ {2} \right] \\ \Rightarrow \quad v _ {m} = \frac {v _ {b}}{2} \\ \text {Further,} \\ \frac {1}{2} M (v _ {m} + 1) ^ {2} = \frac {1}{2} \left(\frac {M}{2}\right) v _ {b} ^ {2} \\ \Rightarrow \quad v _ {m} + 1 = \frac {v _ {b}}{\sqrt {2}} \\ \text {Solving (1) \& (2), we get} \\ v _ {b} = 2 (\sqrt {2} + 1) = 4.8 2 \mathrm{ms} ^ {- 1} \\ v _ {m} = \sqrt {2} + 1 = 2.4 1 \mathrm{ms} ^ {- 1} $
Hence, the correct answer is (C).

Acceleration $= a = \frac{v_0}{t_0}$
$\begin{array}{r l} & P _ {\mathrm{inst}} = F v _ {t} = (m a) v _ {t} = m \left(\frac {v _ {0}}{t _ {0}}\right) (a t) \\ \Rightarrow & P _ {\mathrm{inst}} = m \left(\frac {v _ {0}}{t _ {0}}\right) ^ {2} t \end{array}$
Hence, the correct answer is (D).

In the initial position, the length of the stretched spring is
$L = \frac {h}{\cos 3 7 ^ {\circ}} = \frac {5 h}{4} = 1. 2 5 h$
So, extension in the spring is
x = L - h = 0.2 5 h
PE stored in the spring is
$U = \frac {1}{2} k x ^ {2} = \frac {1}{2} k (0. 2 5 h) ^ {2} = \frac {k h ^ {2}}{3 2}$
When the spring becomes vertical, it acquires natural length and hence the PE gets converted into kinetic energy of the ring
$\Rightarrow \frac {1}{2} m v ^ {2} = \frac {k h ^ {2}}{3 2}$
$\Rightarrow v = \frac {h}{4} \sqrt {\frac {k}{m}}$
Hence, the correct answer is (D).

When the particle at $x_{1}$ is displaced slightly towards left or right, the force is in the same direction. Hence, the equilibrium is unstable, whereas at $x_{2}$ the force is directed opposite to displacement.
Hence, the correct answer is (B).

Since, $a=\left(\frac{m_{2}-m_{1}}{m_{2}+m_{1}}\right)g$ and $v^{2}-u^{2}=2as$
$\Rightarrow v ^ {2} - 0 ^ {2} = 2 \left(\frac {m _ {2} - m _ {1}}{m _ {2} + m _ {1}}\right) g h$
$\Rightarrow v = \sqrt {2 \left(\frac {m _ {2} - m _ {1}}{m _ {2} + m _ {1}}\right) g h}$
Hence, the correct answer is (D).

Acceleration down the plane is $a = g(\sin \theta - \mu \cos \theta)$ velocity of block at time $t$ is $v = at = gt(\sin \theta - \mu \cos \theta)$ Force of friction is $f = \mu mg \cos \theta$
So, rate of work done by force of friction will be
$P = f v = \mu m g ^ {2} t \cos \theta (\sin \theta - \mu \cos \theta)$
Hence, the correct answer is (C).

Let velocity of block at C be v
By Law of Conservation of Energy, we get
$\frac {1}{2} m v _ {0} ^ {2} + m g r (1 - \cos \theta) = \frac {1}{2} m v ^ {2}$
$\Rightarrow v = \sqrt {v _ {0} ^ {2} + 2 g r (1 - \cos \theta)}$
When block leaves the surface, we have
$N = 0$
$\Rightarrow m g \cos \theta = \frac {m v ^ {2}}{r}$
$\Rightarrow r g \cos \theta = v _ {0} ^ {2} + 2 g r (1 - \cos \theta)$
$\Rightarrow 3 r g \cos \theta = \frac {r g}{4} + 2 r g$
$\Rightarrow 3 \cos \theta = \frac {9}{4}$
$\Rightarrow \cos \theta = \frac {3}{4}$
$\Rightarrow \quad \theta = \cos^ {- 1} \left(\frac {3}{4}\right)$
Hence, the correct answer is (B).

x = Elongation in spring due to mass 10 kg
$x = \frac {m g}{k} = \frac {1 0 \times 1 0}{1 0 0} = 1 \mathrm{m}$
$W _ {F} = \frac {1}{2} \times 1 0 0 \times [ (3) ^ {2} - (1) ^ {2} ] - 1 0 \times 1 0 \times 2 = 2 0 0 \mathrm{J}$
Hence, the correct answer is (A).