Question

A 2kg object is released from rest from a height of 10m above the ground. Calculate
(i) the initial potential energy at a moment of release;
(ii) the kinetic energy at the moment it reaches 4m above the ground;
(iii) the speed of the object just before impact with the ground.

Answer 1

Answer:

i) 196J ii) 78.3J iii) 14m/s

Explanation:

i) Formula for gravitational potential energy is

Eg = mgh

Eg = (2)(9.8)(10)

Eg = 196J

ii) Formula for kinetic energy is

Ek = (mv^2)/2

We need to find the velocity first by using one of the key equations of accelerated motion

v^2 = vi^2+2aΔd

v = √vi+2aΔd

v = √2(9.8)(4)

v = 8.85 m/s

Ek = ((2)(8.85^2))/2

Ek = 78.3J

iii)

v = √vi+2aΔd

v = √2aΔd

v = √2(9.8)(10)

v = 14 m/s