Bismark A.
asked • 06/24/16A 60 kg block slides along the top of a 100 kg block with an acceleration of 2.0 m/s2 when a horizo.Find the coefficient of kinetic friction between the blocks.
A 60 kg block slides along the top of a 100 kg block with an acceleration of 2.0 m/s2 when a horizontal force F of 315 N is applied. The 100 kg block sits on a horizontal frictionless surface, but there is friction between the two blocks.
Find the coefficient of kinetic friction between the blocks.
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1 Expert Answer
Joshua Psalms T. answered • 06/24/16
Tutor
5
(5)
Civil EIT, Former College Professor of Mathematics (in Asia)
My solution is:
Horizontal Force - Frictional Force = ma (since there it's hard to draw the force diagram here)
315 - (Friction Constant)*(Normal of 60kg block) = (Mass of 60kg block)*(accelaration)
315 - FC*(60*9.81) = 60*2
Coefficient of Kinetic Friction = 0.331
*What I don't like here is that we haven't used the 100 kg, that's why I would like to know where the force was applied*
Steven W.
tutor
Joshua is exactly right, if the 315 N force is being applied to the upper block. If it is not, then the only force that could accelerate the upper block is the frictional contact. At that point, for the upper block:
Fnet = μkN, where N = weight of upper block (since the normal force is is holding it up against gravity) = 60kg*(9.8 m/s2) = 588 N.
By Newton's 2nd law, Fnet = ma, so that Fnet = (60 kg)(2 m/s2) = 120 N
Thus, 120 N = μk(588 N) --> μk = 120/588 = 0.20
In this case, we use neither the 100 kg NOR the 315 N force, in which case that would have to constitute information thrown into the problem to confuse you and obscure what you really have to focus on.
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07/05/16
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Joshua Psalms T.
06/24/16