Video: Understanding How the Work Required to Move a Body Varies with Friction

You are in a room in a basement with a smooth concrete floor and a nice rug. The rug is 3 m wide and 4 m long. You have to push a very heavy box from one corner of the rug to its opposite corner. The magnitude of friction between the box and the rug is 55 N, but the magnitude of friction between the box and the concrete floor is only 40 N. Will you do more work against friction going around the floor or across the rug? How much extra work would it take?

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Video Transcript

You are in a room in a basement with a smooth concrete floor and a nice rug. The rug is three meters wide and four meters long. You have to push a very heavy box from one corner of the rug to its opposite corner. The magnitude of friction between the box and the rug is 55 newtons, but the magnitude of friction between the box and the concrete floor is only 40 newtons. Will you do more work against friction going around the floor or across the rug? How much extra work would it take?

This is an exercise involving non-conservative forces. And to start out, let’s draw a diagram. In this situation, we have a very heavy crate on the corner of a three-by-four-meter rug. We want to move the crate to the opposite corner. And we’re told that if we move the crate straight across the rug, the force of friction is 55 newtons. While if we choose instead to move the crate on the smooth concrete floor around the rug, the overall force of friction is 40 newtons. We want to figure out which of these two paths will take more work.

Recalling that work is equal to force times distance, we can write that the work required to move the crate across the rug equals 𝐹 sub π‘Ÿ, given as 55 newtons, times the distance across the rug the crate moves. Since the rug is a rectangle, that distance 𝑑 sub π‘Ÿ is equal to the square root of three meters squared plus four meters squared, or five meters. Plugging in these values and calculating π‘Š sub π‘Ÿ, it’s equal to 275 joules.

Next, we wanna calculate the work done if we move the crate across the floor instead of the rug. This is equal to 𝐹 sub 𝑓, which is 40 newtons, times 𝑑 sub 𝑓, which is the distance the crate would move. This distance, skirting the edge of the rug, is equal to four plus three, or seven meters. Plugging in for these two values, when we calculate π‘Š sub 𝑓, we find it’s 280 joules. So, comparing the work done on the rug to the work done on the floor, we see that more work is required to move the crate across the floor by five joules.

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