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A crate of mass 90 kg rests on the floor. How much work is required to move the crate at a constant speed 6 m along the floor against a friction force of 180 N?

Sagot :

LammettHashidn

If the crate is moving at a constant speed, then there is zero net force on the crate, and hence no work is required.

More specifically, by Newton's second law the net horizontal force on the crate is

F - 180 N = 0

where F is the magnitude of the applied force. The right side is zero since constant speed = no acceleration. Then F = 180 N.

Now,

• work done by F = (180 N) (6 m) = 1080 J

• work done by friction = (-180 N) (6 m) = -1080 J

so the total work W is

W = 1080 J - 1080 J = 0 J

Alternatively, by the work-energy theorem, the total work done on the crate is equal to the change in its kinetic energy. But constant speed means initial and final velocities are equal; if v is this speed, then

W = 1/2 (90 kg) v² - 1/2 (90 kg) v² = 0 J

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