(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?

2324 J

Explanation:

The formula for work is:

where is the force applied, and is the distance moved, in this case

and we need to find

Since the crate is not moving up or down, we conclude that the normal force must be equal to the weight of the object:

where is the normal force and is the weight, which is: , where g is the gravitational acceleration and is the mass .

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Thus the normal force is:

Now, the force due to the friction is defined as:

where is 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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no

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