Question

A 56.0 kg mass is sliding down a frictionless plane at a speed of 4.20 m/s. A force of 55.0N parallel to the plane acts on the mass to pull it up the plane. When the force has acted over a distance of 6.80 m, the mass has a speed of 5.40 m/s. What change in gravitational potential energy occurred while the force acted on the mass?

Answer 1

Answer on Question #43398-Physics-Mechanics-Kinematics-Dynamics

A 56.0 kg mass is sliding down a frictionless plane at a speed of 4.20 m/s. A force of 55.0N parallel to the plane acts on the mass to pull it up the plane. When the force has acted over a distance of 6.80 m, the mass has a speed of 5.40 m/s. What change in gravitational potential energy occurred while the force acted on the mass?

Solution

According to the Work Energy Theorem:

$$W = \Delta E _ {\text {g r a v i t a t i o n a l}} + \Delta K,$$

where $W = Fd$ is the work done by the action of force of 55.0N, $\Delta E_{\text{gravitational}}$ is the change in gravitational potential energy, $\Delta K = \frac{m}{2} (v_f^2 - v_i^2)$ is the change in kinetic energy.

The change in gravitational potential energy is

$$\begin{array}{l} \Delta E _ {\text {g r a v i t a t i o n a l}} = W - \Delta K = F d - \frac {m}{2} \left(v _ {f} ^ {2} - v _ {i} ^ {2}\right) = 5 5. 0 \mathrm {N} \cdot 6. 8 0 \mathrm {m} - \frac {5 6 . 0 \mathrm {k g}}{2} \left(\left(5. 4 0 \frac {\mathrm {m}}{\mathrm {s}}\right) ^ {2} - \left(4. 2 0 \frac {\mathrm {m}}{\mathrm {s}}\right) ^ {2}\right) \\ = 5 1. 4 \mathrm {J}. \\ \end{array}$$

Answer: 51.4 J.

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