Question 3) A forklift slowly lifts a 400 kg crate from the floor to a height of

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Question 3) A forklift slowly lifts a 400 kg crate from the floor to a height of 2.00 meters above the floor Lifting slowly means that the acceleration needed to start and stop the lift is negligible a) Find the work done by the forklift when the crate is lifted b) The same crate is raised h = 2.00 meters off the floor by sliding it slowly up an inclined plane as shown below. Assume there is no friction. The mass m2 drops down each time a crate is lifted and after each lift, the fork lift must return m2 to its starting position. How much work is done by each of the forces on each of the objects as the crate slides up the ramp? How much work is done by the forklift to lift m2 back up to its starting position? For each work term that you calculated, state whether it would increase, decrease, or remain the same if there was a small amount of friction for the object sliding on the ramp String Pulley and pulley clamp Container (or cart) with masses inside m, is the total sliding or rolling mass Clamp Weight holder and weights m2 is the total h hanging mass Stand m1 End Stop Clamp Table Ramp

Explanation / Answer

(a) According to Work-Energy Theorem, the amount of work done would be the change in the potential energy of the crate which is given by mgh

mgh = 400*9.81*2 = 7848 J

(b) The total work done is the total change in potential energy.

For mass m1, the total change in potential energy = m1g(2) (Here change in height is 2m)

So the forces acting on m1 would have done a work 2m1g

Similarly for mass m2, the total change in potential energy = m2g(2) (Here also the change in height is 2m)

On considering friction, The work done by forces acting on mass m1 would increase and the work done by forces acting on mass m2 would remain same (Here friction acts as a dissipative source of energy)

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