What is the coefficient of friction between your block and the floor? SHOW ALL WORK. What I know: t=1.06s d=1.87m m=0.132kg initial velocity of the block=3.53 m/s acceleration of the block, as it slid to rest=-3.33 m/s^2 force of friction acting on the block during its slide=-0.44N
Draw a free body diagram, but its not necessary here. To find the force of friction, use F = ma. You know a and m. Force of friction is u*Normal force. Normal force will be equal to the weight, mg, since i'm assuming it's a flat surface
Is it -0.44N=u*(0.132)(9.81) and then I solve for u?
Yup!
Oh okay! Um Part 2 says: If your block slid across a surface with a coefficient of kinetic friction of 0.25, how far would it have slid, assuming that it started with the same initial velocity from #1 (3.53 m/s)?
Use F = ma to find a (use the same method, umg to find friction force F) then use one of the motion equations to find the distance, like \[\Large v_f^2 = v_i^2 + 2a \Delta x\]
I don't get it. :/
Find acceleration, using F=ma. F is the force of friction, umg. Same thing as above, with u=0.25
umg = ma which you can solve for a
would it be 3.33 m/s^2?
umg = ma cancel off m's a = ug = 0.25 (9.81)
oh yeah I got it. I put in the wrong number. :P
so it's 2.4525 m/s^2
Now use the equation above... or some other equation you prefer, using a, to find distance travelled initial v= 3.53 m/s a = 2.4525 m/s^2
-1.44m?
Using the \[\Large v_f^2 = v_i^2 + 2a \Delta x\] vf = 0. vi=3.53. a=2.4525. Solve for the distance delta x
did I do it wrong? I thought I did that. umm.... maybe I'll try again.
Make sure you square the velocity.
oh. -2.54m?
yep :) but it should be positive. It's in the same direction as velocity. But its no big deal as long as you know the direction.
Thanks! I always mess up the calculations when I do them without a calculator.
no prob. Good job!
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