Finding the coefficient of kinetic friction?

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)?

xanthym

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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xelda

I'm still confused. :x

xanthym

Let's answer the questions in order:
#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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Doc Al

Just to add to xanthym's comments:
Apply Newton's 2nd law to find the net force on the object. Then you can figure out what the kinetic friction force must be.

Jameson

F = ma solves for the net amount of force on the object.

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


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

xelda

Thanks for the help! I got the right answer after reading all of your hints, but I'll have to think about it some more to make sense of it.