A force of constant magnitude pushes a box up a vertical surface as shown in the

Question

A force of constant magnitude pushes a box up a vertical surface as shown in the figure. The box moves at a constant speed. If the mass of the box is 3.9 kg, it is pushed 2.8 m vertically upward, the coefficient of friction is 0.35 and the angle is 30.0°, determine the following.

(a) work done on the box by

J (±)

(b) work done on the box by the force of gravity
J (±)

(c) work done on the box by the normal force
J (±)

(d) increase in gravitational potential energy of the system of the box and earth as the box moves up the wall
J

Explanation / Answer

he net force vertically = F*sin30 – 3.9*9.8 - F*cos30 * 0.35

= F*(0.5-0.3031) – 38.22 = 0 (no acceleration)

F = 38.22/(0.5-0.3031) = 194.12N

F = 194.12N

(a) work done on the box by vector F = 194.12*sin30*2.8 = 271.76J

(b) work done on the box by the force of gravity = -3.9*9.8*2.8 = -107 J

(c) work done on the box by the normal force = 0J

(d) increase in gravitational potential energy of the box as it moves up the wall 107J

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