Question

A 200. kg object is pushed 12.0 m to the top of an incline to a height of 6.0 m. If the force applied along the incline is 3000. N, what is the potential energy of the object when it is at the top of the incline with respect to the bottom of the incline?

Answer 1

Approximately
`1.2 * 10^(4)\; {\rm J}` (assuming that
`g = 9.81\; {\rm N \cdot kg^(-1)}`.)

Step-by-step explanation:

The strength of the gravitational field near the surface of the earth is approximately constant:
`g = 9.81\; {\rm N \cdot kg^(-1)}`.

The change in the gravitational potential energy (
`{\rm GPE}`) of an object near the surface of the earth is proportional to the change in the height of this object. If the height of an object of mass
`m` increased by
`\Delta h`, the
`{\rm GPE}` of that object would have increased by
`m\, g\, \Delta h`.

In this question, the height of this object increased by
`\Delta h = 6.0\; {\rm m}`. The mass of this object is
`m = 200\; {\rm kg}`. Thus, the
`{\rm GPE}` of this object would have increased by:

`\begin{aligned}& (\text{Change in GPE}) \\ =\; & m\, g\, \Delta h \\ =\; & 200\; {\rm kg} * 9.81\; {\rm N \cdot kg^(-1)} * 6.0\; {\rm m} \\ \approx\; & 1.2 * 10^(4)\; {\rm J}\end{aligned}`.

(Note that
`1\; {\rm N \cdot m} = 1\; {\rm J}`.)