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?

Question

Felicity Kane

Physics

12 July, 03:59

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?

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Answers (1)

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Faith Macias · Answer 1

a) 1272J

b) 900J

Explanation:

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