You push a 2.0kg box with a constant horizontal force of 50N across a floor 5m.

Question

You push a 2.0kg box with a constant horizontal force of 50N across a floor 5m. of kinetic friction between the box and the floor of 0.3. Is the system Conservative or Non-Conservative? If there is a coefficient of kinetic friction between the box and the floor of 0.3. What is the total work done on the box? If the box starts from rest, how fast will it be moving at the end of the 5m floor. If the box start moving at 2m/s how fast will be moving, after 5m, if the force you push the box with is equal to the force of friction?

Explanation / Answer

(a) System is conservative. Because there is always conservation of momentum and conservation of energy.

(b) Work done on the box, W = F*d = 50*5 = 250 J

(c) Acceleration, a = (F - u*mg) / m = (50 - 0.3*2*9.8) / 2 = 22.06 m/s^2

So, the requisite velocity, v = sqrt(2*22.06*5) = 14.8 m/s

(d) In this case, the force is just equal to neutralize the frictional force. So, there will not be any acceleration.

So, the velocity will be constant and equal to 2 m/s.

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