Work, Power & Energy

In the first case, speed imparted to the marble by the spring is
$\begin{array}{r l} & {\frac {1}{2} k d ^ {2} = \frac {1}{2} m u ^ {2}} \\ {\Rightarrow} & {u = \sqrt {\frac {k}{m}} d} \end{array}$
Since horizontal range of marble is uT, where T is the time of flight. So, we get
$R - r = u T = \left(\sqrt {\frac {k}{m}} d\right) T\tag{... (1}$
Let the required compression in the spring for landing the ball in the box be $x'$ , then
$R = \left(\sqrt {\frac {k}{m}} x ^ {\prime}\right) T\tag{... (2}$
Please note that time of flight is independent of the velocity of the ball and therefore same in both the cases. From (1) and (2), we get
${\frac {R}{R - r}} = {\frac {x ^ {\prime}}{d}}$
$\Rightarrow x ^ {\prime} = \frac {R d}{R - r}$
Hence, the correct answer is (D).

Since $W_{nc} = \Delta U + \Delta K = mgh - (-mgh) + 0 = 2mgh$
Hence, the correct answer is (B).

According to Work Energy Theorem, $W = \Delta K$ , so kinetic energy of block at x = x is
$K = W = \int_ {0} ^ {x} F d x = \int_ {0} ^ {x} (4 - x ^ {2}) d x$
$\Rightarrow K = 4 x - \frac {x ^ {3}}{3}$
For K to be maximum $\frac{dK}{dx}=0$
$\begin{array}{r l} \Rightarrow & 4 - x ^ {2} = 0 \\ \Rightarrow & x = \pm 2 \mathrm{m} \end{array}$
At $x = +2\mathrm{m}$ , we have $\frac{d^2K}{dx^2}$ to be negative
So, kinetic energy (K) is maximum at $x = +2 \, m$
$\Rightarrow K _ {\max} = (4) (2) - \frac {(2) ^ {3}}{3} = \frac {1 6}{3} \mathrm{J} \approx 5. 3 3 \mathrm{J}$
Hence, the correct answer is (C).

Kinetic energy of particle at x = x is
$K = 4 x - \frac {x ^ {3}}{3}$
At maximum x-displacement i.e., at point of reversal of motion, we have
$\begin{array}{r l} & v = 0, \mathrm{so} K = 0 \\ \Rightarrow & x = 2 \sqrt {3} \mathrm{m} \end{array}$
Hence, the correct answer is (A).

Velocity will be vertical at position A as shown in Figure. At this position, tangential acceleration is due to weight. Therefore,
$m a _ {t} = m g$
$\Rightarrow a _ {t} = g$
As the particle is just able to complete the circle, speed at position A is
$v = \sqrt {3 g r}$
Therefore, centripetal acceleration is
$a _ {c} = \frac {v ^ {2}}{r} = \frac {3 g r}{r} = 3 g$
Hence, total acceleration of the particle is
$\begin{array}{r l} & a _ {\mathrm{total}} = \sqrt {a _ {c} ^ {2} + a _ {t} ^ {2}} = \sqrt {(3 g) ^ {2} + g ^ {2}} \\ \Rightarrow & a _ {\mathrm{total}} = \sqrt {1 0} \mathrm{g} \end{array}$
Hence, the correct answer is (C).

Since,
$\begin{array}{r l} & \vec {F} = - \vec {\nabla} U \\ \Rightarrow & \vec {F} = - \lambda (\hat {i} + \hat {j}) \\ \text { Further, } & \Delta \vec {r} = \vec {r} _ {f} - \vec {r} _ {i} \\ \Rightarrow & \Delta \vec {r} = (2 \hat {i} + 3 \hat {j}) - (\hat {i} + \hat {j}) = \hat {i} + 2 \hat {j} \\ \text { So, } & W = \vec {F} \cdot \Delta \vec {r} = - \lambda - 2 \lambda \\ \Rightarrow & W = - 3 \lambda \\ & \text { Hence, the correct answer is (B). } \end{array}$

Work done = Area under F-s graph
Area = $36 + (-6) = 30$ J
Hence, the correct answer is (B).

Let the maximum extension in the spring be $x_{max} = x_{m}$ , then by Applying Work-Energy Theorem from the frame attached to point A, we get
$(m a) x _ {m} - \frac {1}{2} k x _ {m} ^ {2} = 0$
$\Rightarrow x _ {m} = \frac {2 m a}{k}$
Hence, the correct answer is (C).

$\vec{v} = \alpha \hat{i} +\beta t\hat{j}$
$\Rightarrow \vec {a} = \frac {d \vec {v}}{d t} = \beta \hat {j}$
So, total acceleration $|\vec{a}| = \beta$
The speed, $v = \sqrt{\alpha^{2} + \beta^{2}t^{2}}$
So, tangential acceleration, $a_{T}$ is
$a _ {T} = \frac {d v}{d t} = \frac {1}{2} \Big (\alpha^ {2} + \beta^ {2} t ^ {2} \Big) ^ {1 / 2} \Big (2 \beta^ {2} t \Big)$
At $t = \frac{\sqrt{3}\alpha}{\beta}$
Tangential acceleration, $a_{T}=\frac{\sqrt{3}\beta}{2}$
and normal acceleration, $a_{N}=\sqrt{\beta^{2}-\frac{3\beta^{2}}{4}}=\frac{\beta}{2}$ Hence, the correct answer is (A).

$\sin \theta = \frac {5}{1 0} = \frac {1}{2}$
$\Rightarrow \theta = 3 0 ^ {\circ} $
$W = f s \cos 1 8 0 = - f s $
$\Rightarrow W = - (\mu m g \cos 3 0) s $
$\Rightarrow W = - (0.1) (0.5) (1 0) \left(\frac {\sqrt {3}}{2}\right) (1 0 + 1 0) $
$\Rightarrow W = - 5 \sqrt {3} J $
Hence, the correct answer is (D).

Let x be the extension in the spring when 2 kg block leaves the contact with ground. Then,
$\begin{array}{r l} & k x = 2 \mathrm{g} \\ \Rightarrow & x = \frac {2 g}{k} = \frac {2 \times 1 0}{4 0} = \frac {1}{2} \mathrm{m} \end{array}$
Now by Law of Conservation of Energy, we have
$\begin{array}{r l r} & & m g x = \frac {1}{2} k x ^ {2} + \frac {1}{2} m v ^ {2} \qquad \{\mathrm{where} m = 5 \mathrm{kg} \} \\ & \Rightarrow & v = \sqrt {2 g x - \frac {k x ^ {2}}{m}} \\ & \Rightarrow & v = \sqrt {2 \times 1 0 \times \frac {1}{2} - \frac {4 0}{4 \times 5}} = 2 \sqrt {2} \mathrm{ms} ^ {- 1} \end{array}$
Hence, the correct answer is (B).

The ball rotates in a circle around the nail after it goes past the point B.
At point B, velocity of the ball is $v = \sqrt{2gL}$ ...(1)
Since the ball just completes the vertical circle about the nail, its speed at the lowest point should be
$v = \sqrt {5 g r} = \sqrt {5 g (L - d)}$....(2)
Equating (1) and (2), we get
$\sqrt {2 g L} = \sqrt {5 g (L - d)}$
$\Rightarrow 2 L = 5 L - 5 d$
$\Rightarrow d = 0.6 L$
Hence, the correct answer is (A).

$\begin{array}{r l} & {- f s = \frac {1}{2} m v ^ {2} - \frac {1}{2} m u ^ {2}} \\ {\Rightarrow} & {- f \bigg (\frac {2 0}{1 0 0} \bigg) = \frac {1}{2} \bigg (\frac {6 0}{1 0 0 0} \bigg) \big [ (3 0 0) ^ {2} - (6 0 0) ^ {2} \big ]} \\ {\Rightarrow} & {f \bigg (\frac {2 0}{1 0 0} \bigg) = \frac {1}{2} \bigg (\frac {6 0}{1 0 0 0} \bigg) \big (1 0 ^ {4} \big) (3 6 - 9)} \\ {\Rightarrow} & {f = 1.5 \times 1 0 ^ {3} \times 2 7 = 4 0.5 \times 1 0 ^ {3} \mathrm{N} = 4 0.5 \mathrm{kN}} \end{array}$
Hence, the correct answer is (A).

When the particle looses its contact with the sphere, then we know that
$\cos \theta = \frac {2}{3}$
$\Rightarrow \sin \theta = \frac {\sqrt {5}}{3}$
So, tangential acceleration at this position is
$a _ {T} = g \sin \theta = \frac {\sqrt {5} g}{3}$
Hence, the correct answer is (B).

The cable does not work on the bob. So, the bob's energy is conserved
$\Rightarrow \frac {1}{2} m v _ {0} ^ {2} = m g h.....(1)$
Where $h = \ell (1 - \cos \theta)$
... (2)
From equations (1) and (2), we get
$v _ {0} = 7 \mathrm{ms} ^ {- 1}$
Hence, the correct answer is (B).

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).