Question

A block of mass 10 kg slides down an inclined plane that has an angle of 30. If the inclined plane has no friction and the block starts at a height of 2 m, how much kinetic energy does the block have when it reaches the bottom? Acceleration due to gravity is g = 9.8 m/s2.

Answer 1

Answer:

E = 196 joules

Explanation:

It is given that,

Mass of the block, m = 10 kg

At the top, the block will have only potential energy which is given by :

$P=mgh$

At the bottom of the inclined plane, it will have only kinetic energy which is given by :

$E=\dfrac{1}{2}mv^2$

Applying the conservation of energy as :

$E=\dfrac{1}{2}mv^2=mgh$

$E=10\ kg\times 9.8\ m/s^2\times 2\ m$

E = 196 joules

So, the block will have 196 joules of kinetic energy when it reaches the bottom. Hence, this is the required solution.

Answer 2

No friction present means: Ek = Ep

So Ek = mgh = 10 * 9.8 * 2 = 196 J