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(II) How much work did the movers do (horizontally) pushing a 46.0-kg crate 10.3 m across a rough floor without acceleration, if the effective coefficient of friction was 0.50?
(II) How much work did the movers do (horizontally) pushing a 46.0-kg crate 10.3 m across a rough floor without acceleration, if the effective coefficient of friction was 0.50?
Answer:2324 JExplanation:The formula forworkis:whereis the force applied, andis the distance moved, in this caseand we need to findSince the crate is not moving up or down, we conclude that thenormal force must be equal to the weightof the object:whereis the normal force andis the weight, which is:, where g is the gravitational accelerationandis the mass.---------Thus the normal force is:Now, the force due to the friction is defined as:whereis the coefficient of friction,So,for the crate to move, the force applied must be equal to the frictional force:And now that we know the force we can calculate the work:substituting known values:...
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