Coefficient of kinetic friction homework

[KrazeeNym]

Please help me solve this problem

A girl coasts down a hill on a sled, reaching level ground at the bottom with a speed of 7.0 m/s. The coefficient of kinetic friction between the sled's runners and the hard, icy snow is 0.050, and the girl and sled together weigh 645 N. How far does the sled travel on the level ground before coming to rest?

Thanks in advance

 

[OlderDan]

Please help me solve this problem

A girl coasts down a hill on a sled, reaching level ground at the bottom with a speed of 7.0 m/s. The coefficient of kinetic friction between the sled's runners and the hard, icy snow is 0.050, and the girl and sled together weigh 645 N. How far does the sled travel on the level ground before coming to rest?

Thanks in advance

We will help you if you show us your attempt to solve a problem. We will not do a problem for you.

 

[kinza]

I'm having trouble with the same problem. So far I have attempted to calculate the Ff.

mu=.050 or Ff/645 N
Ff=32.3 N

I also plan to use the formula, vf^2=vi^2+2ax but I need the acceleration. So I need to calculate the sum of all forces which is, Ff fight? The mass is 66 kg. So using f=ma, is the acceleration .49 m/s^2? Using the above equation, is the distance traveled 48 m? Thanks so much guys.

 

[OlderDan]

I'm having trouble with the same problem. So far I have attempted to calculate the Ff.

mu=.050 or Ff/645 N
Ff=32.3 N

I also plan to use the formula, vf^2=vi^2+2ax but I need the acceleration. So I need to calculate the sum of all forces which is, Ff fight? The mass is 66 kg. So using f=ma, is the acceleration .49 m/s^2? Using the above equation, is the distance traveled 48 m? Thanks so much guys.

Your approach is correct. Your answer is a bit off. You should check your computation. Also, if you take the direction of motion as positive, the acceleration is actually negative.

 

[kinza]

Thanks for responding! The answer is a little off? I thought it would be a lot off. I really have no idea where I went wrong. Are the formulas for mu and vf correct? Also, does 32.3 N seem plausible for the friction?

 

[OlderDan]

Thanks for responding! The answer is a little off? I thought it would be a lot off. I really have no idea where I went wrong. Are the formulas for mu and vf correct? Also, does 32.3 N seem plausible for the friction?

The friction is F = μN and N is the weight in this case (horizontal ground with no vertical forces other than gravity and the normal) You have that part done perfectly. The acceleration comes from F = -μN = ma, so a = -μN/m = -μNg/W = -.49m/s², which you also have found correctly except for the sign. Your only problem is the computation based on vf² = vi² + 2ax. It is the correct relationship to use, but your answer for x is not consistent with the values that go into the equation.

 

[kinza]

Hmm...I tried again and got 50 m. Is that right?

 

[OlderDan]

Hmm...I tried again and got 50 m. Is that right?

Looks good to me.

 

[kinza]

Thanks a million! (really)

 

[tidalwav1990]

I am currently working on the same problem and I am majorly stuck. I know the answer is 50 but i am not sure how to get it. I am using Vf^2=Vi^2 + 2AX.
Vf=7
Vi=0
A=-.05

I believe those are the right numbers but when i plug them in I get -490 not 50. Can anyone help me, i'd really appreciate it.

 

[tidalwav1990]

nm. I finally get it now I just had the wrong acceleration. sorry to anyone that attempted to help me or bothered reading this.