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A 200 N force with direction drawn above is pulling a crate along rough ground.

There is a coefficient of kinetic friction of 0.200 between the crate and the ground.

The crate has a weight of 50.0 N. at a angle of 20 degress

How much work is done by the frictional force if the crate is moved 3.00 m to the right?

User Atiana
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1 Answer

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To solve this problem, we need to first calculate the net force acting on the crate. We can do this using the following equation:

Net force = Force applied - Force of friction - Weight of crate

In this case, the force applied is 200 N to the right, the weight of the crate is 50 N downwards at an angle of 20 degrees, and the force of friction can be calculated using the coefficient of kinetic friction and the normal force (which is equal in magnitude to the weight of the crate, since the crate is on a horizontal surface):

Force of friction = coefficient of friction * normal force

= 0.200 * 50.0 N

= 10.0 N to the left (since friction opposes motion)

Net force = 200 N - 10.0 N - 50.0 N sin(20)

= 132.4 N to the right (rounded to one decimal place)

Now that we know the net force, we can use the work-energy principle to calculate the work done by the frictional force:

Work done by friction = Force of friction * distance moved

= 10.0 N * 3.00 m

= 30.0 J

Therefore, the work done by the frictional force is 30.0 J.

User Sampgun
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