A 30.0 kg packing crate in a warehouse is pushed to the loading dock by a worker who applies a horizontal force. The coefficient of kinetic friction between the crate and the floor is 0.20. The loading dock is 15.0 m southwest of the initial position of the crate. (a) If the crate is pushed 10.6 m south and then 10.6 m west, what is the total work done on the crate by friction? (b) If the crate is pushed along a straight-line path to the dock, so that it travels 15.0 m southwest, what is the work done on the crate by friction?

In order to calculate work, we need the magnitude of the friction and the distance the crate moved. For part a) the distance is the sum of 10.6m + 10.6m and for part b) it is 15m. So now we just need the magnitude of the friction.

We know that Fr = μ*N where N = m*g, so:

Fr = μ * m * g Replacing the values, we get Fr = 0.2*30*10 = 60N

With this magnitude we can now calculate the work:

W1 = Fr * d1 + Fr * d2 = 60*10.6 + 60*10.6 = 636J + 636J = 1272 J

And now through the straight line:

W2 = Fr * d = 900J