A 4.0 kg box of fruit slides 8.0 m down a ramp, inclined at 30.° from the horizontal. If the box slides at a constant velocity of 5.0 m/s, the work done by gravity is
A) 0.16 kJ.
B) -0.16 kJ.
C) -78. J.
D) zero.
E) +78. J.

Question

Quizplus · Accepted Answer

The work done by gravity is equal to the change in potential energy of the box. The change in potential energy is given by:$$\Delta U = mgh$$where:* m is the mass of the box (4.0 kg)* g is the acceleration due to gravity (9.8 m/s^2)* h is the change in height of the box (8.0 m * sin(30°) = 4.0 m)Plugging in these values, we get:$$\Delta U = (4.0 kg)(9.8 m/s^2)(4.0 m) = 156.8 J$$Since the box is sliding at a constant velocity, the work done by gravity is equal to the work done by friction. Therefore, the work done by gravity is -156.8 J, which is closest to choice A) -0.16 kJ.

A 40 Kg Box of Fruit Slides 8