It is given to us that

Θ = 30°

F = 46 N

F_{W} = 25 N

d = 3 m

µ_{k} = 0.2•

Sorry that cannot provide the picture, but imagine the following forces applied to the body :

**F**_{W} - weight of the body directed vertically down

**F** - external force parallel to the incline and in upward direction

**N** - normal reaction force perpendicular to the surface

**f** - kinetic friction force directed parallel to the incline and in downward direction.

In all cases we will be using the formula work W = force times displacement in the direction of force (here where cos of the angle between the force and displacement is hidden).

a) For gravitational force Wa = F_{W}•d•cosα , here α is the angle between vectors **F**_{W} and **d. **From the picture we see that α = 120°. If we put the numbers, we will get that the work by gravitational force is - 75 J.

b) For external force W_{b} = F•d•cosβ, here β is the angle between vectors **F** and **d**, hence 0°. Please put the numbers by yourself.

c) For friction force W_{c} = f•d•cosγ, here γ is the angle between vectors **f** and **d**, hence 180°.

The magnitude of f = µ_{k}N_{ } and N = F_{W}•cosΘ, then f = µ_{k}•F_{W}•cosΘ. I hope you can do the rest for this part of the problem yourself. If you did everything right, you will see that the work of friction force is negative.

d) for normal force W_{d }= N•d•cosδ, where δ is the angle between vectors **N** and **d**, hence 90°. Because the cos90° = 0, the work of Normal force is 0.

I hope that you understood the problem and have a good holiday!