A 5 kg box drops a distance of 10 m to the ground. If 70% of the initial potential energy goes into increasing the internal energy of the box, determine the magnitude of the increase.

Explanation:

From the given information:

The initial PE = m×g×h

= 5 kg × 9.81 m/s² × 10 m

= 490.5 J

The change in Potential energy P.E of the box is:

ΔP.E =

ΔP.E = 0 -

ΔP.E =

If we take a look at conservation of total energy for determining the change in the internal energy of the box;

this can be re-written as:

Here, K.E = 0

Also, 70% goes into raising the internal energy for the box;

Thus,

ΔU = 343.35  J

Thus, the magnitude of the increase is = 343.35 J

answer; equivalent to a one(1) horsepower motor;