Question

A 0.5 kg ball rolls down a frictionless inclined plane. The height of the inclined plane is 0.5m. What is the kinetic energy of the ball when it reaches the bottom of the inclined plane?

Answer 1

Answer:

2.45Joules

Explanation:

since kinetic energy is K.E.=1/2m$v^{2}$

lost in potential energy is gain of kinetic energy

mgh=1/2mv^2

9.8*2*0.5=v^2

$v^{2}$=9.8

v=$\sqrt{9.8}$ =3.13m/s

now k.e. = 0.5*m*$v^{2}$ =0.5*0.5*9.8 =2.45J