work by friction

Question

A 3.1 kg block is slid along the floor through different paths as shown above. The co-efficient of kinetic friction between the block and the floor is 0.23.

1)

What is the work down by friction when the block moves from point A to point B along back 3?

W_Fric-3 =

2)

What is the work done by friction if the block moves in a closed loop, starting at point A along path 2 and then back along path 3?

W_Fric-23 =

3)

What is the work done by friction if the blcok moves in a closed loop, starting at point A along path 1 and then back along path 2?

W_Fric-12 =

Answer 1

Work can be expressed as force times the displacement that the force acts through. For friction, though, things are a bit different. Friction is a nonconservative force, meaning that friction could act along a path that just goes in a circle and ends up at the starting point, and will NOT do zero work, unlike conservative forces such as gravity and spring forces, which would do 0 work. Instead, the work done by friction (and other nonconservative forces) depends on the path taken.

Thus, the work done by kinetic friction is NOT force * displacement, but rather force * distance traveled.

Then, we have, w = -f · d, where f is the friction force and d is the distance traveled. Note that there is a negative sign because kinetic friction here always acts against the direction of motion of the block.

We also have that kinetic friction is f = µ_kF_n, where µ_k is the coefficient of kinetic friction, and F_n is the normal force, which, in this case, is equal to the block's weight, mg (assuming flat ground).

Putting it together, we have,

w = -µ_kmgd.

1) w = -µ_kmgd = -(0.23)(3.1 kg)(9.8 m/s^2)(1.0 m + 3.0 m + 3.0 m) = -49 joules

2) w = -µ_kmgd = -(0.23)(3.1 kg)(9.8 m/s^2)(2.0 m + 2.0 m + 1.0 m + 1.0 m + 3.0 m + 3.0 m) = -84 joules

3) w = -µ_kmgd = -(0.23)(3.1 kg)(9.8 m/s^2)(4.0 m + 4.0 m + 1.0 m + 1.0 m + 1.0 m + 1.0 m + 2.0 m + 2.0 m) = -110 joules

Hope that helped!