A 10-kg box is at the top of a 5-m plane inclined at 37º with the horizontal.The box starts from rest and slides down the plane.The coefficient of kinetic friction is 0.20.The magnitude of the change in the potential energy of the box is
A)30 kJ
B)0.10 kJ
C)0.29 kJ
D)0.46 kJ
E)0.39 kJ

Question

Quizplus · Accepted Answer

First, we need to find the gravitational potential energy at the top of the incline:
$$U_{g,i} = mgh = (10\,\mathrm{kg})\left(9.81\,\mathrm{\frac{m}{s^2}}\right)(5\,\mathrm{m}\sin 37^\circ) \approx 235.6\,\mathrm{J}$$
Then, we need to find the work done by the kinetic friction:
$$W_{fr} = f_k d = \mu_k N d = \mu_k mgd\cos 37^\circ = (0.20)(10\,\mathrm{kg})(9.81\,\mathrm{\frac{m}{s^2}})(5\,\mathrm{m})\cos 37^\circ \approx 19.6\,\mathrm{J}$$
where $N$ is the normal force and $d$ is the length of the incline. Since the box starts from rest, all of the potential energy is converted to kinetic energy when it reaches the bottom of the incline. Finally, we can find the change in potential energy:
$$\Delta U = U_{g,i} - W_{fr} = 235.6\,\mathrm{J} - 19.6\,\mathrm{J} = 216\,\mathrm{J} \approx 0.29\,\mathrm{kJ}$$ 
Therefore, the correct choice is (C).

A 10-Kg Box Is at the Top of a 5-M