Question

A 50-gram ball is released from rest 80 m above the surface of the Earth. During the fall to the Earth, the total thermal energy of the ball and the air in the system increases by 15 J. Just before it hits the surface its speed is

Answer 1

Answer:

Speed of ball just before it hit the surface is 31.62 m/s .

Explanation:

Given :

Mass of ball , m = 50 g = 0.05 kg .

Height from which it falls , h = 80 m .

Thermal energy , E = 15 J .

Now , Initial energy of the system is :

$E_i=\dfrac{mv^2}{2}+mgh$

Here , initial velocity is zero .

Therefore , $E_i=mgh=40\ J$

Now , final energy of the system :

$E_f=\dfrac{mv^2}{2}+mg(0)+15\\\\E_f=\dfrac{0.05\times v^2}{2}+15$

Since , no external force is applied .

Therefore , total energy of the system will be constant .

By conservation of energy :

$E_i=E_f\\40=\dfrac{0.05v^2}{2}+15\\\\25=\dfrac{0.05v^2}{2}\\\\v=31.62\ m/s$

Therefore , speed of ball just before it hit the surface is 31.62 m/s .