Solveeit Logo
Login

Question

Question: A ball of mass, \(m = 0.2\,kg\) starts from rest at a height \({y_0} = 15\,m\) above the surface of ...

A ball of mass, m=0.2 kgm = 0.2\,kg starts from rest at a height y0=15 m{y_0} = 15\,m above the surface of the earth and falls down to a height yf=0.5 m{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.

Explanation

Solution

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 kgm = 0.2\,kg
Since it starts from rest so, initial velocity of particle is, u=0msec⁡−1u = 0m{\sec ^{ - 1}}
Also let the final velocity of the ball be vv.
Since the initial height of ball is, y0=15 m{y_0} = 15\,m
So the potential energy stored in ball at such height is, U1=mgy0{U_1} = mg{y_0}
Putting values we get,
U1=(0.2 kg)(9.8 msec⁡−2)(15 m){U_1} = (0.2\,kg)(9.8\,m{\sec ^{ - 2}})(15\,m)
⇒U1=29.4J\Rightarrow {U_1} = 29.4J
After reaching at height yf=0.5 m{y_f} = 0.5\,m its new potential energy would be,
U2=mgyf{U_2} = mg{y_f}
Putting values we get,
U2=(0.2 kg)(9.8 msec⁡−2)(0.5 m){U_2} = (0.2\,kg)(9.8\,m{\sec ^{ - 2}})(0.5\,m)
⇒U2=0.98J\Rightarrow {U_2} = 0.98J
Since potential energy decrease, so amount of decrease is given by,
ΔU=U1−U2\Delta U = {U_1} - {U_2}
⇒ΔU=(29.4−0.98)J\Rightarrow \Delta U = (29.4 - 0.98)J
⇒ΔU=28.42 J\Rightarrow \Delta U = 28.42\,J

Let the change in kinetic energy is, ΔK\Delta K. Since the mechanical energy always remains constant, so decrease in potential energy is equal to increase in kinetic energy that is, ΔK=ΔU\Delta K = \Delta U
Putting values we get, ΔK=28.42J\Delta K = 28.42J
So change in kinetic energy is given by, ΔK=28.42J\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,
d⃗=(yf−yi)j^\vec d = ({y_f} - {y_i})\hat j
⇒d⃗=(0.5−15)j^\Rightarrow \vec d = (0.5 - 15)\hat j
⇒d⃗=−14.95j^\Rightarrow \vec d = - 14.95\hat j
So work done is defined as,
W=F⃗.d⃗W = \vec F.\vec d

Putting values we get,
W=(−mgj^).(−14.95j^)W = ( - mg{\kern 1pt} \hat j).( - 14.95\hat j)
⇒W=14.95mg(j^.j^)\Rightarrow W = 14.95mg(\hat j.\hat j)
⇒W=14.95mg\Rightarrow W = 14.95mg
⇒W=(14.95 m)(0.2 kg)(9.8 msec⁡−2)\Rightarrow W = (14.95\,m)(0.2\,kg)(9.8\,m{\sec ^{ - 2}})
⇒W=29.302J\Rightarrow W = 29.302J
Since from work energy theorem we know,
W=ΔKW = \Delta K
⇒W=12mv2−12mu2\Rightarrow W = \dfrac{1}{2}m{v^2} - \dfrac{1}{2}m{u^2}
Putting values we get,
29.302=12(0.2)v2−029.302 = \dfrac{1}{2}(0.2){v^2} - 0
v2=293.02 m2sec⁡−2{v^2} = 293.02\,{m^2}{\sec ^{ - 2}}

So final velocity of ball is given as, v=17.12 msec⁡−1v = 17.12\,m{\sec ^{ - 1}} and change in kinetic energy is given by, ΔK=28.42 J\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.