## Mastering Physics Solutions: Pushing Too Hard

Pushing Too Hard

Part A = F1 < F2 <= 2F1
Part B = 1/2F1 <= F3 < F1

Solution Below:

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?

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

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?

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

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### 2 Responses to Mastering Physics Solutions: Pushing Too Hard

1. Anonymous says:

This is incorrect. The first answer is F1<F2< or = 2F1 and the second answer is 1/2F1< or = F3<F1

• Mastering Physics Solutions says:

Whoops! You’re absolutely right, thank you for the correction!