Question

A baggage handler at an airport applies a constant horizontal force with magnitude F₁ to push a box, of mass m, across a rough horizontal surface with a very small constant acceleration a.

Part A

The baggage handler now pushes a second box, identical to the first, so that it accelerates at a rate of 2a. How does the magnitude of the force F₂ that the handler applies to this box compare to the magnitude of the force F₁ applied to the first box?

Part B

Now assume that the baggage handler pushes a third box of mass m/2 so that it accelerates at a rate of 2a. How does the magnitude of the force F₃ that the handler applies to this box compare to the magnitude of the force F₁ applied to the first box?

Answer 1

Part A Answer

This question is confusing. Mastering Physics doesn’t give you all the information you need to solve it correctly, but the answer it wants is F1 < F2 <= 2F1.

• 0 <= F2 < 1/2F1
• 1/2F1 <= F2 < F1
• F2 = F1
• F1 < F2 <= 2F1
• F2 > 2F1

F1 < F2 <= 2F1

Part B Answer

Remember F = ma. Since the acceleration has doubled but the mass has been cut in half, F = (1/2m)(2a) = ma. This is equal to F₁

• 0 <= F3 < 1/2F1
• 1/2F1 <= F3 < F1
• F3 = F1
• F1 < F3 < 2F1
• F3 > 2F1

1/2F1 <= F3 < F1