# Drowning in Physics

#### PurposeDriven

If anyone could help me on a physics question, I would sincerely appreciate it. I have already done it, but somehow I feel it's wrong . If someone could check it and tell me what I'm doing wrong if it is wrong, that would be great!!

1) A 400g object with an initial speed of 80cm/s slides along a horizontal desktop against a frictional force of .7N.
A) How far will it slide before stopping?
B) What is the coefficient of friction between the object and the
table top?

So what I have so far is M=400g, Initial Velocity=80cm/s, Force of kinetic friction is .7N, Final Velocity is 0, and average velocity is 40cm/s. I know I first want to solve for d.

I used F=Ma ( .7N=400g(a) ) and found that acceleration is 1.75m/sec^2
I then used t=V(final)-V(initial)/a to solve for time. I converted my initial velocity into meters/sec. (-.8/1.75= -.4571428571) This makes sense that it's negative since it's deaccelerating.
Then I plugged that inThen I used V(av)=d/t to solve for d. (.4m/s*.4571428571=.1828571429) I dropped the negative b/c distance can't be negative. (Can I do that?) So my answer for A is .182857429m

Now for B I used the equation Mu sub k(coefficient of kinetic friction)=Force of Kinetic friction/Normal Force. So basically, Mu-sub-k=.7/(1.4*9.8) and ended up with .1785714286e-4.

Thank you so much!!!!

#### Doc Al

Mentor
Originally posted by PurposeDriven
I used F=Ma ( .7N=400g(a) ) and found that acceleration is 1.75m/sec^2
I then used t=V(final)-V(initial)/a to solve for time. I converted my initial velocity into meters/sec. (-.8/1.75= -.4571428571) This makes sense that it's negative since it's deaccelerating.
Be more careful with your signs. The acceleration is negative, but the time better be positive!
Then I plugged that inThen I used V(av)=d/t to solve for d. (.4m/s*.4571428571=.1828571429) I dropped the negative b/c distance can't be negative. (Can I do that?) So my answer for A is .182857429m
Your general method (and your answer) is fine. If you are careful with signs, nothing will need to be "dropped". Also, don't use so many significant figures in your answer: three is plenty.

Of course, you did much uneeded work. If you are able to use energy methods, that's the way to go:
Workdone by friction = &Delta;KE
Now for B I used the equation Mu sub k(coefficient of kinetic friction)=Force of Kinetic friction/Normal Force. So basically, Mu-sub-k=.7/(1.4*9.8) and ended up with .1785714286e-4.
Your method is correct, but you used the wrong value for mass (0.4, not 1.4). Redo the calculation and you should get the right answer.

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