Question 401263: A heavy, 94-kg crate is pushed a distance of 5.2 m at a speed of 0.40 m/s on the floor of a
warehouse. The coefficient of the kinetic friction is 0.25.
a) Determine the magnitude of the horizontal force needed to move the crate.
b) Find out the work done in moving the crate.
c) What is the power involved in moving the crate? (g = 9,8 m/s2)
Answer by richard1234(7193)   (Show Source):
You can put this solution on YOUR website! a) Use the equation  , where  is the coefficient of kinetic friction, and N is the normal force. Substituting, we get
 (Newtons)
The net force is zero (net force meaning sum of all the forces), since the crate is pushed at a constant velocity (hence no acceleration), so a 230.3 N force is required.
b) Work = force*distance. The work done by the person is given by  (joules). However, the net work done on the block is still zero because the net force on the block is zero.
c) Power is the rate that work is done, so Power = Work/time. We know the work done by the person is 1198 J. We need to find the amount of time taken.
We know that distance = rate*time, so 5.2 m = (0.40 m/s)*time. Solving, we get time = 13 seconds.
Therefore, Power = (1198 J)/(13 s) = 92.1 Watts.
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