A 6.0-kg block is released from rest 80 m above the ground. When it has fallen 60 m its kinetic energy is approximately:

Question

Quizplus · Accepted Answer

The potential energy of the block at the height of 80 m can be calculated as:

PE = mgh = 6.0 kg x 9.81 m/s^2 x 80 m = 4708.8 J

When the block falls to a height of 60 m, its potential energy decreases and its kinetic energy increases. The work done by the gravitational force is converted into kinetic energy. Let's assume the velocity of the block when it has fallen 60 m is v.

The work done by the gravitational force is:

W = Fd = mgh - mgh' = ΔPE

where h' = 60 m and h = 80 m.

So, ΔPE = PE - PE' = 4708.8 J - mgh' = 4708.8 J - 6.0 kg x 9.81 m/s^2 x 60 m = 1765.28 J

This work done is equal to the change in kinetic energy:

ΔKE = KE - KE' = 1/2 mv^2 - 1/2 mv0^2 = 1/2 mv^2

where v0 = 0 (initially at rest).

Substituting ΔKE and solving for v gives:

ΔKE = 1/2 mv^2 = 1765.28 J

v = sqrt(2ΔKE/m) = sqrt(2 x 1765.28 J / 6.0 kg) = 28.87 m/s

The kinetic energy of the block when it has fallen 60 m can be calculated as:

KE = 1/2 mv^2 = 1/2 x 6.0 kg x (28.87 m/s)^2 = 2983.5 J

Thus, the kinetic energy of the block when it has fallen 60 m is approximately 3500 J (closest to B).

A 60-Kg Block Is Released from Rest 80 M Above the the Ground