$A$ raindrop of mass $1.00 \, g$ falling from a height of $1 \, km$ hits the ground with a speed of $50 \, m s^{-1}$. Calculate
$(a)$ the loss of $PE$ of the drop
$(b)$ the gain in $KE$ of the drop
$(c)$ Is the gain in $KE$ equal to loss of $PE$? If not,why?
Take $g = 10 \, m s^{-2}$.

$A$ small block of mass $M$ moves on a frictionless surface of an inclined plane,as shown in the figure. The angle of the incline suddenly changes from $60^{\circ}$ to $30^{\circ}$ at point $B$. The block is initially at rest at $A$. Assume that collisions between the block and the incline are totally inelastic $\left(g=10 \ m/s^2\right)$.
$1.$ The speed of the block at point $B$ immediately after it strikes the second incline is
$(A) \sqrt{60} \ m/s$ $(B) \sqrt{45} \ m/s$ $(C) \sqrt{30} \ m/s$ $(D) \sqrt{15} \ m/s$
$2.$ The speed of the block at point $C$,immediately before it leaves the second incline is
$(A) \sqrt{120} \ m/s$ $(B) \sqrt{105} \ m/s$ $(C) \sqrt{90} \ m/s$ $(D) \sqrt{75} \ m/s$
$3.$ If the collision between the block and the incline is completely elastic,then the vertical (upward) component of the velocity of the block at point $B$,immediately after it strikes the second incline is
$(A) \sqrt{30} \ m/s$ $(B) \sqrt{15} \ m/s$ $(C) 0$ $(D) -\sqrt{15} \ m/s$
Give the answers for questions $1, 2,$ and $3.$

$A$ block of mass $5 \ kg$ is suspended by a light thread from an elevator. The elevator is accelerating upward with uniform acceleration $2 \ m/s^2$. The work done by tension on the block during the first $3 \ s$ is $(u = 0)$. (in $J$)

$A$ bullet of mass $30 \,g$ moving with $700 \,ms^{-1}$ collides with a block of mass $4 \,kg$ hanging by a string of length $0.4 \,m$. After the collision, the block rises to a height of $0.2 \,m$. Find the velocity of the bullet when it emerges from the block. (in $\,ms^{-1}$)

The displacement $x$ of a particle moving in one dimension under the action of a constant force is related to the time $t$ by the equation $t = \sqrt{x} + 3$,where $x$ is in meters and $t$ is in seconds. The work done by the force in the first $6$ seconds is.....$J$

$A$ person trying to lose weight by burning fat lifts a mass of $10 \ kg$ up to a height of $1 \ m$, $1000$ times. Assume that the potential energy lost each time he lowers the mass is dissipated. How much fat will he use up considering the work done only when the weight is lifted up? Fat supplies $3.8 \times 10^7 \ J$ of energy per $kg$, which is converted to mechanical energy with a $20\%$ efficiency rate. Take $g = 9.8 \ m/s^2$.