Question

A 6-kg block slides down an incline with a 5-meter vertical drop over an 8-meter horizontal distance. If the block starts from rest and friction is negligible, then what is its kinetic energy at the bottom?

Answer 1

To solve this problem we will apply the concepts related to energy conservation. We know that potential energy is transformed into kinetic and vice versa energy. Since the energy accumulated in the upper part is conserved as potential energy, when the object is thrown all that energy will be converted into kinetic energy. Therefore we will have the following relation,

$KE = PE$

$KE = mgh$

Here,

m = mass

g = Gravitational acceleration

h = Height

Replacing,

$KE = (6)(9.8)(5)$

$KE = 294J$

Therefore the kinetic energy at the bottom is 294J