Question

A 20.0-kg rock is dropped and hits the ground at a speed of 90.0 m/s. Calculate the rock’s gravitational potential energy before it was dropped. (Ignore the effects of friction.)

Answer 1

Explanation :

It is given that,

Mass of the rock, m = 20 kg

Initially the rock is at rest, u = 0

Final velocity of the rock is, v = 90 m/s

Gravitational potential energy is given by :

$U=mgh$............(1)

h is the height.

Using third equation of motion :

$v^2-u^2=2ah$

or

$h=\dfrac{v^2}{2g}$

Put the value of h in equation (1)

$U=mg\times \dfrac{v^2}{2g}$

$U=\dfrac{mv^2}{2}$

$U=\dfrac{20\ kg\times (90\ m/s)^2}{2}$

$U=81000\ J$

or

U = 81 kJ

Hence, this is the required solution.

Answer 2

Answer:

GPE=81000J or 81kJ

Explanation

Potential Energy = mgh = 20 x 9.8 x ? 

To find H use one of the equation of motion 

= [(90)^2 - 0 ] / 2(9.8) 

Potential Energy = mgh = 20 x 9.8 x 8100 /2(9.8) = 81000 J