What is the work Ww done on the box by the weight of the box?

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.

Nov 13, 2017

OK, first we need to find the force pulling the box down the plane, then figure out the work done.

Explanation:

The friction is given by ${F}_{r} = \mu . N$ where N is the normal reaction, 1.7N

So ${F}_{r} = 0.3 \times 1.7 = 0.51$ N this points against the motion, so acts up the plane.

The force acting down the plane is given by the component of gravity acting parallel to the plane, ${F}_{a} = m . g . \sin \theta$

${F}_{a} = 2 \times \sin {30}^{\circ}$ = 1 N

So the net force, $F = {F}_{a} - {F}_{r}$

$F = 0.49$ N

Work done, W = F.s

So $W = 0.49 \times 1.8$

Thus W = 0.882 or 0.88 Nm to 2 s.f.