A box of weight w=2.0N accelerates down a rough plane that is inclined at an angle ϕ=30∘ above the horizontal, as shown (Figure 6). The normal force acting on the box has a magnitude n=1.7N, the coefficient of kinetic friction between the box and the plane is μk=0.30, and the displacement d⃗ of the box is 1.8 m down the inclined plane. What is the work Ww done on the box by the weight of the box?

Question

Sanaa Larson

Physics

19 September, 03:09

A box of weight w=2.0N accelerates down a rough plane that is inclined at an angle ϕ=30∘ above the horizontal, as shown (Figure 6). The normal force acting on the box has a magnitude n=1.7N, the coefficient of kinetic friction between the box and the plane is μk=0.30, and the displacement d⃗ of the box is 1.8 m down the inclined plane. What is the work Ww done on the box by the weight of the box?

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Answers (1)

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Sloane Sparks · Answer 1

First, we must calculate the resultant force on the box.

Force due to weight:

weight x sin (Ф)

= 2 x sin (30)

= 1 N

Frictional force =

normal force x coefficient of friction

= 1.7 x 0.3

= 0.51 N

Resultant force = 1 - 0.51

= 0.49 N

Work = Force x displacement in direction of force

= 0.49 x 1.8

= 0.882 Joules