Work, Power & Energy

Ans. (C)
Applying WET
$- m g (h + R) = \frac {1}{2} m v _ {B} ^ {2} - \frac {1}{2} m v _ {0} ^ {2}$
Considering motion from B to C
$v _ {B} \times \sqrt {\frac {2 R}{g}} = d \Rightarrow v _ {B} = d \sqrt {\frac {g}{2 R}}$
Putting in equation (i) we get option (C).

Ans. (A) Particle should not loose contact any where on the circular track. Considering $N = 0$ at highest point. $\frac{mv^2}{R} = mg \Rightarrow d = \sqrt{2} R$

Ans. (B) $v_{0}^{2} - v_{A}^{2} = 2gh$ $F_{\mathrm{A}} = \frac{mv_A^2}{R} = \frac{m(v_0^2 - 2gh)}{R}$

$k = \frac{1}{2} mv^2$ $k = \frac{1}{2} \times 5 \times 400 = 5 \times 200 = 1000J$ $\frac{\Delta k}{k} = \frac{\Delta m}{m} + \frac{2\Delta v}{v} = \frac{0.5}{5} + \frac{2 \times 0.4}{20}$ $\Delta k = 1000\left(\frac{1}{10} + \frac{4}{100}\right) = 1000\left(\frac{10 + 4}{100}\right) = 140J.$

Work done = Area enclosed by F - x graph $= \frac{1}{2} \times (3 + 6) \times 3 = 13.5 \, J .$

$U_{i} = 0$ and $U_{f} = \frac{1}{2} k(2x_{0})^{2}$
$\Rightarrow U _ {f} = 2 k x _ {0} ^ {2}$
Since we know that, $W_{C} = -\Delta U$

If W be the work done by the spring on earth mass, then
$ W _ {C} = 2 W = 0 - 2 k x _ {0} ^ {2}$
$\Rightarrow W = - k x _ {0} ^ {2} $
Hence, the correct answer is (D).

Force needed to pull the block down by a distance y from equilibrium position is
F = k (x + y) - m g = k y
Therefore, work done to extend the spring is
$W = \frac {1}{2} k y ^ {2}$
At equilibrium, mg = kx
$\Rightarrow k = \frac {m g}{x} = \frac {2 \times 1 0}{0.0 6} = \frac {1 0 0 0}{3} \mathrm{Nm} ^ {- 1}$
When pulled down by additional distance 10 cm, external work done $W=\frac{1}{2}\times k\times(0.1)^{2}$
$\Rightarrow W = \frac {1}{2} \times \frac {1 0 0 0}{3} \times 0. 0 1 = 1. 6 7 \mathrm{J}$
Hence, the correct answer is (D).

By Law of Conservation of Energy
$\begin{array}{r l} & m g (h + x) = \frac {1}{2} k x ^ {2} \\ \Rightarrow & x = \frac {m g + \sqrt {m ^ {2} g ^ {2} + 2 m g h k}}{k} \end{array}$
Hence, the correct answer is (A).

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).