Work, Power & Energy

From Work Energy Theorem $W = \Delta KE$
$\Rightarrow W = K _ {f} - K _ {i} = \frac {1}{2} m \Big (v _ {f} ^ {2} - v _ {i} ^ {2} \Big)$
$\Rightarrow W = \frac {1}{2} (2) (4 0 0 - 1 0 0) = 3 0 0 \mathrm{J}$
Hence, the correct answer is (A).

$P _ {\mathrm{inst}} = \vec {F} \cdot \vec {v} = 5 0 - 3 0 + 1 2 0 = 1 4 0 \mathrm{W}$
Hence, the correct answer is (C).

At extreme position A, since the ball is at rest so, net acceleration is equal to the tangential acceleration.
$\Rightarrow a _ {A} = g \sin \theta$
At lowermost position B, net acceleration is centripetal acceleration, i.e.,
$\begin{array}{r l} & a _ {B} = \frac {v ^ {2}}{L} \text {where} v = \sqrt {2 g L (1 - \cos \theta)} \\ \Rightarrow & a _ {B} = 2 g (1 - \cos \theta) \\ \text {Since,} a _ {A} = a _ {B} \\ \Rightarrow & g \sin \theta = 2 g (1 - \cos \theta) \\ \Rightarrow & 2 g \sin \left(\frac {\theta}{2}\right) \cos \left(\frac {\theta}{2}\right) = 2 g \left(1 - \left(1 - 2 \sin^ {2} \frac {\theta}{2}\right)\right) \end{array}$
Since $\sin \theta = 2\sin \left(\frac{\theta}{2}\right)\cos \left(\frac{\theta}{2}\right)$ and
$1 - \cos \theta = 2 \sin^ {2} \left(\frac {\theta}{2}\right)$
$\begin{array}{l l} \Rightarrow & 2 g \sin \left(\frac {\theta}{2}\right) \cos \left(\frac {\theta}{2}\right) = 2 g \times 2 \sin^ {2} \left(\frac {\theta}{2}\right) \\ \Rightarrow & \tan \frac {\theta}{2} = \frac {1}{2} \\ \Rightarrow & \frac {\theta}{2} = 2 6. 5 ^ {\circ} \\ \Rightarrow & \theta = 5 3 ^ {\circ} \end{array}$

$v = \frac {d x}{d t} = 3 - 8 t + 3 t ^ {2}$
Initial velocity (velocity at t=0) is $u=3\ ms^{-1}$ . Final velocity (velocity at t=4s) is $v=19\ ms^{-1}$
$\begin{array}{l} \text {Since,} W = \frac {1}{2} m (v ^ {2} - u ^ {2}) = \frac {1}{2} \left(\frac {3}{1 0 0 0}\right) (3 6 1 - 9) \\ \Rightarrow \quad W = \frac {5 2 8}{1 0 0 0} \mathrm{J} = 5 2 8 \mathrm{mJ} \end{array}$
Hence, the correct answer is (B).

The displacement of block A in ground frame towards right is s (as shown in figure).
$\Rightarrow s = 0.5 - 0.3 = 0.2 \mathrm{m}$
Kinetic friction acting on block A acting towards right is
$f _ {k} = \mu m g = 0.2 \times 4 5 \times 1 0 = 9 0 \mathrm{N}$
So, work done by friction force in the ground frame is
$W = f s \cos (0 ^ {\circ}) = 9 0 \times 0.2 \times 1 = 1 8 \mathrm{J}$
Hence, the correct answer is (B).

$\begin{array}{l} {a = \left(\frac {m _ {2} - \mu m _ {1}}{m _ {1} + m _ {2}}\right) g} \\ {\Rightarrow \quad a = \left[ \frac {4 - (0.2) (1 0)}{4 + 1 0} \right] 9.8} \end{array}$
$\Rightarrow a = 1.4 \mathrm{ms} ^ {- 2}$
$\mathrm{Since}, v ^ {2} - 0 ^ {2} = 2 a s$
$\Rightarrow v ^ {2} = 2 (1.4) (4)$
$\Rightarrow \quad v ^ {2} = 1 1.2$
$\Rightarrow v \cong 3.5 \mathrm{ms} ^ {- 1}$
Hence, the correct answer is (D).

Since, $\vec{F}$ is a constant force, so work done will not depend on path. So,
$\begin{array}{l l} & W = \int_ {(0, 0, 0)} ^ {(2, 2, 2)} \vec {F} \cdot d \vec {\ell} \\ \Rightarrow & W = \int_ {(0, 0, 0)} ^ {(2, 2, 2)} (2 \hat {i} + 5 \hat {j} + \hat {k}) \cdot (d x \hat {i} + d y \hat {j} + d z \hat {k}) \\ \Rightarrow & W = \int_ {(0, 0, 0)} ^ {(2, 2, 2)} (2 d x + 5 d y + d z) \\ \Rightarrow & W = (2 x + 5 y + z) | _ {(0, 0, 0)} ^ {(2, 2, 2)} = 1 6 J \end{array}$
Hence, the correct answer is (D).

Given force is a constant force and work done is path independent under a constant force. Hence, $W_{1} = W_{2}$ . Hence, the correct answer is (A).

$\text { Since } \Delta U = - \int \vec {F} \cdot d \vec {l}$
$\Rightarrow U - U _ {0} = \int 4 0 (y d x + x d y) = 4 0 \int d (x y)$
$\Rightarrow U - U _ {0} = 4 0 x y $
$\Rightarrow U = U _ {0} + 4 0 x y$
At the mid point (C) of the rod, the particle is in equilibrium.

Also, we must note that from A till mid point C, the force is towards A, so the particle faces opposition from force to reach the mid point C. However after C i.e., between C and B, the force is directed towards B. This makes us conclude that if we give the particle a sufficient speed just to reach the mid point C, then after crossing C it will automatically reach B. Hence by Law of Conservation of Energy applied between A and C we have
$\begin{array}{l l} & \binom {\text {Loss in KE}} {\text {of particle}} = \binom {\text {Gain in PE}} {\text {of particle}} \\ \Rightarrow & \frac {1}{2} m v _ {0} ^ {2} = 4 0 \bigg (\frac {2}{2} \bigg) \bigg (\frac {2}{2} \bigg) - 4 0 (0) (2) \\ \Rightarrow & \frac {1}{2} (5) v _ {0} ^ {2} = 4 0 \\ \Rightarrow & v _ {0} = \sqrt {\frac {8 0}{5}} = \sqrt {1 6} = 4 \mathrm{ms} ^ {- 1} \end{array}$
Hence, the correct answer is (D).

Method-I:
$d W = \vec {F} \cdot d \vec {\ell} = 2 d x + 3 d y$
Along the path $3y + kx = 5$ , we have
$3 d y + k d x = 0$.....(i)
$\begin{array}{l l} \text {Since W = 0 i.e.,} \int d W = 0 \\ \Rightarrow \quad \int 2 d x - k d x = 0 & \left\{\because \text {from (1),} 3 d y = - k d x \right\} \\ \Rightarrow \quad k = 2 \end{array}$
Method-II:
Force is parallel to a line $y = \frac{3}{2} x + c$
The equation of given line can be written as
$y = - \frac {k}{3} x + \frac {5}{3}$
Work done will be zero, when force is perpendicular to the displacement i.e., the above two lines are perpendicular or
$\begin{array}{r l} & m _ {1} m _ {2} = - 1 \\ \Rightarrow & \left(\frac {3}{2}\right) \left(- \frac {k}{3}\right) = - 1 \\ \Rightarrow & k = 2 \end{array}$
Hence, the correct answer is (A).

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).