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?
1 answer:
Answer:
2.45Joules
Explanation:
since kinetic energy is K.E.=1/2m
lost in potential energy is gain of kinetic energy
mgh=1/2mv^2
9.8*2*0.5=v^2
=9.8
v=
=3.13m/s
now k.e. = 0.5*m* =0.5*0.5*9.8 =2.45J
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