## Finding the coefficient of kinetic friction?

A force of 45.7 N is required to start a 5.0-kg box moving across a horizontal concrete floor. (a) What is the coefficient of static friction between the box and the floor? (b) If the force continues at the same value, the box accelerates at 0.54 m/s^2. What is the coefficient of kinetic friction?

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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Recognitions:
 Quote by xelda A force of 45.7 N is required to start a 5.0-kg box moving across a horizontal concrete floor. (a) What is the coefficient of static friction between the box and the floor? (b) If the force continues at the same value, the box accelerates at 0.54 m/s^2. What is the coefficient of kinetic friction? 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 expect?? How does that compare with the given 0.54 m/s^2 acceleration?? What would account for the difference?? (Hint: F=ma)

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 I'm still confused. :x

Recognitions:

## Finding the coefficient of kinetic friction?

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?? How does that compare with the given 0.54 m/s^2 acceleration?? What would account for the difference?? (Hint: F=ma)
#1:
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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