A 10-kg box is at the top of a 5.0-m plane inclined at 37^o with the horizontal. The box starts from rest and slides down the plane. The coefficient of kinetic friction is 0.20. The change in the potential energy of the box is

Question

Quizplus · Accepted Answer

The change in potential energy of the box is given by:

ΔPE = mgh

Where m is the mass of the box, g is the acceleration due to gravity, and h is the height of the box above some reference level. In this case, we can take the reference level to be the bottom of the incline, so h = 0.

The initial potential energy of the box is:

PEi = mgh = (10 kg)(9.8 m/s^2)(5.0 m)sin(37°) = 239 J

At the bottom of the incline, the box has a final velocity of:

vf = sqrt(2gh(sinθ - μcosθ))

Where μ is the coefficient of kinetic friction and θ is the angle of the incline. Plugging in the numbers, we get:

vf = sqrt(2(9.8 m/s^2)(5.0 m)(sin37° - 0.20cos37°)) ≈ 5.43 m/s

The final kinetic energy of the box is:

KEf = (1/2)mvf^2 = (1/2)(10 kg)(5.43 m/s)^2 ≈ 147 J

The change in kinetic energy is:

ΔKE = KEf - KEi = 147 J - 0 J = 147 J

Therefore, the change in potential energy is:

ΔPE = -ΔKE = -147 J ≈ -0.15 kJ

Since the change in potential energy is negative, the correct answer is C) -0.29 kJ (rounded to two decimal places).

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