Need Help with solution

Question

A 25N block is being pushed up on a plane inclined at an angle of 30° with the horizontal 46 N force applied parallel to the plane. The block moves 3 m up the inclined plane. How much work is done on the block by:

a.   gravitational force?

b.   Force

c.   Friction force

d.   Normal force

The coefficient of friction µ_k is 0.2 .

Answer 1

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_kN 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!