Question

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.

Answer 1

Answer:

Explanation:

From the given information:

The initial PE $(PE)_i$ = 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 = $P.E_f -P.E_i$

ΔP.E = 0 - $(PE)_i$

ΔP.E = $-P.E_i$

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

$\Delta P.E + \Delta K.E + \Delta U = 0$

$\Delta U = -\Delta P.E - \Delta K.E$

this can be re-written as:

$\Delta U =- (-\Delta P.E_i) - \Delta K.E$

Here, K.E = 0

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

Thus,

$\Delta U =(70\%) \Delta P.E_i-0$

$\Delta U =(0.70) (490.5)$

ΔU = 343.35  J

Thus, the magnitude of the increase is = 343.35 J