Question

A ball of mass, $m = 0.2\,kg$ starts from rest at a height ${y_0} = 15\,m$ above the surface of the earth and falls down to a height ${y_f} = 0.5\,m$ above the surface of earth. What is the change in kinetic energy? Find the final velocity using the work energy theorem.

Answer 1

Hint: In order to solve this question we need to understand work and work energy theorem. Work is defined as a scalar product of force and displacement, it is a scalar quantity as scalar product results in production of only magnitude. Work Energy Theorem states that work done by mechanical force is equal to change in kinetic energy. In this question we would first use that mechanical energy always remains a constant concept then use work energy theorem to calculate final velocity.

Complete step by step answer:
Given the mass of ball is, $m = 0.2\,kg$
Since it starts from rest so, initial velocity of particle is, $u = 0m{\sec ^{ - 1}}$
Also let the final velocity of the ball be $v$.
Since the initial height of ball is, ${y_0} = 15\,m$
So the potential energy stored in ball at such height is, ${U_1} = mg{y_0}$
Putting values we get,
${U_1} = (0.2\,kg)(9.8\,m{\sec ^{ - 2}})(15\,m)$
$\Rightarrow {U_1} = 29.4J$
After reaching at height ${y_f} = 0.5\,m$ its new potential energy would be,
${U_2} = mg{y_f}$
Putting values we get,
${U_2} = (0.2\,kg)(9.8\,m{\sec ^{ - 2}})(0.5\,m)$
$\Rightarrow {U_2} = 0.98J$
Since potential energy decrease, so amount of decrease is given by,
$\Delta U = {U_1} - {U_2}$
$\Rightarrow \Delta U = (29.4 - 0.98)J$
$\Rightarrow \Delta U = 28.42\,J$

Let the change in kinetic energy is, $\Delta K$. Since the mechanical energy always remains constant, so decrease in potential energy is equal to increase in kinetic energy that is, $\Delta K = \Delta U$
Putting values we get, $\Delta K = 28.42J$
So change in kinetic energy is given by, $\Delta K = 28.42J$
Here only the force of gravity is acting on the body given by, $\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\rightharpoonup}$}} {F} = mg( - \hat j)$ it is in downward direction.Here, the negative sign is due to the fact that we have considered that downward direction is negative in nature.And displacement of particle is given as,
$\vec d = ({y_f} - {y_i})\hat j$
$\Rightarrow \vec d = (0.5 - 15)\hat j$
$\Rightarrow \vec d = - 14.95\hat j$
So work done is defined as,
$W = \vec F.\vec d$

Putting values we get,
$W = ( - mg{\kern 1pt} \hat j).( - 14.95\hat j)$
$\Rightarrow W = 14.95mg(\hat j.\hat j)$
$\Rightarrow W = 14.95mg$
$\Rightarrow W = (14.95\,m)(0.2\,kg)(9.8\,m{\sec ^{ - 2}})$
$\Rightarrow W = 29.302J$
Since from work energy theorem we know,
$W = \Delta K$
$\Rightarrow W = \dfrac{1}{2}m{v^2} - \dfrac{1}{2}m{u^2}$
Putting values we get,
$29.302 = \dfrac{1}{2}(0.2){v^2} - 0$
${v^2} = 293.02\,{m^2}{\sec ^{ - 2}}$

So final velocity of ball is given as, $v = 17.12\,m{\sec ^{ - 1}}$ and change in kinetic energy is given by, $\Delta K = 28.42\,J$

Note: It should be remembered that here we have considered the total mechanical energy to remain constant because force of gravity is conservative in nature, also the force is conservative in nature because potential energy is not dependent on time rather it only depends on space. For the case where potential is time dependent then force is no longer conservative.