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Coefficient of friction/ dynamics problem

  • Thread starter lah214
  • Start date
11
0
1. Homework Statement
Lilly, whos mass is 45.0 kg, is ice skating with a constant speed of 7.00 m/s when she hits a rough patch of ice with a coefficient of friction of 0.0800. How long will it take before Lilly coasts to a stop?

m=45 kg
v=7m/s
coefficient of friction=.08

2. Homework Equations
equation 1: f=(coefficient of friction) x Normal Force
equation 2: v=u+at


3. The Attempt at a Solution

I know that the force of friction is 35.2 N. I got this by using equation 1.

I know that Lilly is no longer going at a constant speed once she hits the rough patch of ice, so I know she is not in equilibrium, and therefore there must be some acceleration. In order to find the acceleration though, I must know the net force. I'm having trouble figuring out how to find that.

Can anyone please give me a hint on how to find the net force? Thanks in advance if you can!
 
434
0
If you've found the friction, you've found the net force in the horizontal direction. Keep in mind that the frictional force points opposite to the motion of the skater, and therefore the skater will have a constant deceleration. You know that fk = ma by Newton's Second Law, now you can find the deceleration. Does that help?
 
Last edited:
11
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ahh!! yes, thank you soo much! I always forget that the net force is just friction, because it seems weird thats its going in the opposite direction as the net force. Thank you soo much!
 
11
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it being the skater
 
434
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ahh!! yes, thank you soo much! I always forget that the net force is just friction, because it seems weird thats its going in the opposite direction as the net force. Thank you soo much!
You're welcome. Just to make sure you have a handle on what's going on, remember that the net force is equal to ma. We juse happened to encounter a case where the friction was the only force acting in the horizontal direction. If someone was pushing the skater at the same time, we'd have to account for the friction and the pushing force in the net force equation.
 
11
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Oh, yeah, that's actually what we've been doing in my class lately: the pushing at the same time. It had been some time since we had a problem where the only force was friction, hah, and I spent like 20 minutes trying to figure out how to do it. Anyway, thanks for all the help, I really appreciate it!
 

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