Work, Power & Energy

According to Work-Energy Theorem, work done equals change in kinetic energy, so
$W = F x = \frac {1}{2} m v ^ {2} = \Delta K$
Since for both the cars of different mass, $\Delta K$ will be different, so W is also different.
Hence, the correct answer is (D).

Since $m_{car} < m_{truck}$ , so retardation on the car is more than the truck and hence the car will stop earlier in lesser distance.
Hence, the correct answer is (D).

$F = -\frac{dU}{dx}$
$F = 0$ at points $B$ and $C$ .
Hence, the correct answer is (D).

At the topmost point, the velocity is $\sqrt{gR} (\neq 0)$ , because to just complete the circle, velocity at lowest point is $\sqrt{5gR}$ . At the highest point velocity is zero, however acceleration is non-zero equal to $g$ . Hence, the correct answer is (D).

$\frac {1}{2} m v ^ {2} = m g h _ {1}$
$\Rightarrow h _ {1} = \frac {v ^ {2}}{2 g}$
$\mathrm{Also}, \frac {1}{2} (2 m) v ^ {2} = (2 m) g h _ {2}$
$\Rightarrow h _ {2} = \frac {v ^ {2}}{2 g}$
Hence, the correct answer is (A).

Work done and power developed is zero in uniform circular motion, because $\vec{F} \perp \vec{r}$ . Also $P = \frac{W}{t}$ , so Statement-2 is the correct explanation to Statement-1. Hence, the correct answer is (A).

$W (m g) + W (s p) + W (M a n) = 0$
$\Rightarrow - \Delta U + E + W = 0$
$\Rightarrow \Delta U = E + W$
Work done by spring force is positive when a compressed spring is released or stretched spring released.
Hence, the correct answer is (C).

By Law of Conservation of Momentum, we have
$O = m _ {1} v _ {1} - m _ {2} v _ {2}$
and $K_{1} + K_{2} = E$ the energy released in explosion
$\Rightarrow \frac {K _ {1}}{K _ {2}} = \frac {m _ {2}}{m _ {1}}$
$\Rightarrow P = \sqrt {2 m E}$
Hence, the correct answer is (C).

Statement-1 is false because normal force cannot do any work in the absence of any displacement. The Statement-2 is correct because an external force alone can shift the position of centre of mass.
Hence, the correct answer is (D).

Just before the cord comes in contact with the peg, we have

Hence, the correct answer is (C).

Just after the cord comes in contact with the peg, we have
$\begin{array}{r l} & h = 0.8 \sin (3 0 ^ {\circ}) = 0.4 \\ \Rightarrow & v ^ {2} = 2 g h \\ \text {So,} T _ {2} - m g \sin (3 0 ^ {\circ}) = \frac {m v ^ {2}}{R _ {2}} \\ \Rightarrow & T _ {2} = \frac {m g}{2} + \frac {m (2 g) (0.4)}{0.4} \\ \Rightarrow & T _ {2} = \frac {5 m g}{2} \end{array}$
Hence, the correct answer is (A).

At maximum extension (situation of momentary rest),
$v _ {A} = v _ {B} = z e r o$
By Law of Conservation of Energy
$\begin{array}{l l} & \left( \begin{array}{c} \text {Loss of Gravitational} \\ \text {Potential Energy} \\ \text {of block B} \end{array} \right) = \left( \begin{array}{c} \text {Gain in Elastic} \\ \text {Potential Energy} \\ \text {of the spring} \end{array} \right) \\ \Rightarrow & (2 m) g x _ {\max} = \frac {1}{2} k x _ {\max} ^ {2} \\ \Rightarrow & x _ {\max} = \frac {4 m g}{k} \end{array}$
Hence, the correct answer is (D).

At $x=\frac{x_{max}}{2}=\frac{2mg}{k}$ , we have
$v _ {A} = v _ {B} = v \mathrm{(say)}$
{Because A and B are connected by a rigid string}
Then by Law of Conservation of Energy
$\begin{array}{l l} & \left( \begin{array}{c} \text {Decrease} \\ \text {in GPE} \\ \text {of B} \end{array} \right) = \left( \begin{array}{c} \text {Increase in} \\ \text {EPE of} \\ \text {the spring} \end{array} \right) + \left( \begin{array}{c} \text {Increase} \\ \text {in KE of} \\ \text {both blocks} \end{array} \right) \\ \Rightarrow & (2 m) g x = \frac {1}{2} k x ^ {2} + \frac {1}{2} (m + 2 m) v ^ {2} \\ \Rightarrow & (2 m g) \bigg (\frac {2 m g}{k} \bigg) = \frac {1}{2} k \bigg (\frac {2 m g}{k} \bigg) ^ {2} + \frac {1}{2} (3 m) v ^ {2} \\ \Rightarrow & \frac {8 m ^ {2} g ^ {2}}{k} = \frac {4 m ^ {2} g ^ {2}}{k} + 3 m v ^ {2} \\ \Rightarrow & 3 m v ^ {2} = \frac {4 m ^ {2} g ^ {2}}{k} \\ \Rightarrow & v = 2 g \sqrt {\frac {m}{3 k}} \end{array}$
Hence, the correct answer is (D).

At $x = \frac{x_{\max}}{4} = \frac{mg}{k}$ , we draw the free body diagrams for the blocks A and B, below

Since A and B are connected with a rigid thread, so at any instant $a_{A} = a_{B} = a$ (say)
$\Rightarrow T - k x = m a$
$\Rightarrow T - m g = m a$
... (1)
$\left\{\because x = \frac {m g}{k} \right\}$
Also, $2mg - T = 2ma$
... (2)
Solving these two equations, we get
m g = 3 m a
$\Rightarrow a = \frac {g}{3}$
Hence, the correct answer is (C).

By Law of Conservation of Energy
$\frac {1}{2} m v ^ {2} = m g H - m g h$
$\Rightarrow v = \sqrt {2 g (H - h)}$
Hence, the correct answer is (D).

From concepts of projectile motion, we have
$h = \frac {1}{2} g t ^ {2}$
$\Rightarrow t = \sqrt {\frac {2 h}{g}}$
Hence, the correct answer is (B).

$s = vt$
$\Rightarrow s = \sqrt {2 g (H - h)} \sqrt {\frac {2 h}{g}}$
$\Rightarrow s = 2 \sqrt {h (H - h)}$
... (1)
For s to be MAXIMUM, $s^{2}$ has to be MAXIMUM.
$\frac {d (s ^ {2})}{d h} = 0$
$\Rightarrow \frac {d}{d h} [ 4 h (H - h) ] = 0$
$\Rightarrow H - 2 h = 0$
$\Rightarrow h = \frac {H}{2}$
Hence, the correct answer is (B).

$s_{\max} = 2\sqrt{\left(\frac{H}{2}\right)\left(\frac{H}{2}\right)}$ $\Rightarrow s_{\max} = H$ Hence, the correct answer is (D).

Total height $h = 2 + \frac{u_{B}^{2} \sin^{2} \theta}{2g}$
$\Rightarrow h = 2 + \frac {2 g (h _ {A} - h _ {B}) \sin^ {2} \theta}{2 g}$
$\Rightarrow h = 2 + 4 \sin^ {2} \theta$
We can see that h = 6 m only when $\theta = 90^{\circ}$ Hence, the correct answer is (B).

Since, $h = 2 + 4\sin^{2}\theta$ , so at $\theta = 30^{\circ}$ , h = 3 m Hence, the correct answer is (A).

$dW = \vec{F} \cdot d\vec{\ell}$ $\Rightarrow \quad dW = k\left(x^{2}y^{3}dx + x^{3}y^{2}dy\right)$ $\Rightarrow \quad dW = k\left[y^{3}d\left(\frac{x^{3}}{3}\right) + x^{3}d\left(\frac{y^{3}}{3}\right)\right]$ $\Rightarrow \quad dW = k\left[d\left(\frac{x^{3}y^{3}}{3}\right)\right]$ $\Rightarrow \quad W = \int dW = \frac{k}{3}(x^{2}y^{3})$ $W_{(0,0)\to(a,0)} = \frac{k}{3}(x^{3}y^{3})\bigg|_{(0,0)}^{(a,0)} = \frac{24}{3}(0-0) = zero$
Hence, the correct answer is (A).

$W_{(a,0)\to (a,a)} = \frac{k}{3}\Big(x^3 y^3\Big)\Big|_{(a,0)}^{(a,a)} = \frac{9}{3}\big(a^6 -0\big) = 3a^6$
Hence, the correct answer is (C).

$W _ {(0, 0) \rightarrow (a, a)} = \frac {k}{3} \left(x ^ {3} y ^ {3}\right) \Bigg | _ {(0, 0)} ^ {(a, a)} = \frac {k}{3} \left(a ^ {6} - 0\right) = \frac {k a ^ {6}}{3}$
Hence, the correct answer is (C).

Since we observe that the work done in all the three paths is same i.e., $\frac{ka^{6}}{3}$ . So, work done due to this force is independent of the path followed between $(0, 0)$ to $(a, a)$ . Hence $\vec{F}$ is a conservative force.
Hence, the correct answer is (A).

$W_{mg} = mgh \cos 180^{\circ}$ $\Rightarrow W_{mg} = (2)(10)(2)(-1) = -40 \, \text{J}$ From Work Energy Theorem, we get $W_{mg} + W_{upthrust} = \Delta KE$ $\Rightarrow -40 \, J + W_{upthrust} = 16$ $\Rightarrow W_{upthrust} = 56 \, J$ Hence, the correct answer is (D).