Question

A 13.7-kg box is being pushed from the bottom to the top of a frictionless ramp. When the box is pushed at a constant velocity, the nonconservative pushing force does 58.0 J of work. How much work is done by the pushing force when the box starts from rest at the bottom and reaches the top of the same ramp with a speed of 1.50 m/s?

Answer 1

A 13.7-kg box is being pushed from the bottom to the top of a frictionless ramp. When the box is pushed at a constant velocity, the nonconservative pushing force does 58.0 J of work. How much work is done by the pushing force when the box starts from rest at the bottom and reaches the top of the same ramp with a speed of 1.50 m/s?

Solution:

The system will be the box and the earth. Then the only external force is the pushing force. We have:

work done by external force = change of energy of the system

That is:

work done by pushing force = $\Delta U_{k} + \Delta U_{g}$

(Uk: kinetic energy, Ug: potential energy of gravity)

$$A = \Delta U_{k} + \Delta U_{g}$$ $$\Delta U_{g} = A - \Delta U_{k}$$

in the first part, the box moves with a constant velocity, so there is no change in kinetic energy. Hence, $\Delta U_{k} = 0$

$$\Delta U_{g} = A - 0 = A = 58 \text{ J}$$

Next, we can rewrite the equation ($U_{k_{\text{initial}}}$ is zero since the box starts from rest):

work done by pushing force = $\Delta U_{k} + \Delta U_{g}$

$$A_{\text{push}} = \Delta U_{k} + \Delta U_{g}$$ $$\begin{aligned} A_{\text{push}} &= U_{k_{\text{final}}} - U_{k_{\text{initial}}} + A = \frac{mV^{2}}{2} - 0 + A = \\ &= \frac{13.7kg \cdot \left(1.5 \frac{m}{s}\right)^{2}}{2} + 58 \text{ J} = 73.4 \text{ J} \end{aligned}$$

Answer: work is done by the pushing force is 73.4 J.