- #1

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I've been able to solve part (a) which turned out to be .933. What am I supposed to do with the acceleration to solve for part (b)?

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- Thread starter xelda
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- #1

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I've been able to solve part (a) which turned out to be .933. What am I supposed to do with the acceleration to solve for part (b)?

- #2

xanthym

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xelda said:

I've been able to solve part (a) which turned out to be .933. What am I supposed to do with the acceleration to solve for part (b)?

A 45.7 N force acts on the 5.0 kg box, so you might expect acceleration. How much acceleration would you

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- #3

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I'm still confused. :x

- #4

xanthym

Science Advisor

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Let's answer the questions in order:

A 45.7 N force acts on the 5.0 kg box, so you might expect acceleration. How much acceleration would you expect?? (Consider only the 45.7 N force acting alone.)

(Hint: F=ma)

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#1:A 45.7 N force acts on the 5.0 kg box, so you might expect acceleration. How much acceleration would you expect?? How does that compare with the given 0.54 m/s^2 acceleration?? What would account for the difference?? (Hint: F=ma)

A 45.7 N force acts on the 5.0 kg box, so you might expect acceleration. How much acceleration would you expect?? (Consider only the 45.7 N force acting alone.)

(Hint: F=ma)

~~

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- #5

Doc Al

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Apply Newton's 2nd law to find thexelda said:What am I supposed to do with the acceleration to solve for part (b)?

- #6

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In this problem, you have the Frictional Force going one way and the Applied Force (shown by the acceleration) going the other way.

Ff = (mu) * the Normal Force

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Fnet = Fa + Ff (one of these will be negative)

You know the applied force and you know the net force. Go from there.

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