A block is at rest on a plank whose angle can be varied. The angle is gradually increased from 0 deg. At 32.9°?
A block is at rest on a plank whose angle can be varied. The angle is gradually increased from 0 deg. At 32.9°, the block starts to slide down the incline, traveling 2.80 m down the incline in 2.10 s.
Calculate the kinetic coefficent of friction between the block and the plank.
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Okay, so if you do a force balance on the block in the direction parallel and perpendicular to the incline, you will have:
Perpendicular: N - m*g*cos(th) = 0
Parallel: u*N - m*g*sin(th) = m*a
where u is the coefficient of kinetic friction and N is the normal force.
Combining and eliminating N, we have:
u*m*g*cos(th) - m*g*sin(th) = m*a
Since the LHS is not dependent on time, the acceleration is constant and can be found with the kinematic equation:
D = v0*t + 0.5*a*t^2 with v0 = 0 (starting from rest)
So a = 2*D/t^2
Therefore, if we substitute for a and solve for u, we have:
u = (2*D/t^2 + g*sin(th)) / (g*cos(th))
u = (2*2.80/2.10^2 + 9.81*sin(32.9)) / (9.81*cos(32.9))
u = 0.80
**The answer below is wrong. He plugged in the numbers wrong. This is why I always leave everything in terms of variables until the very end...you are less likely to make a mistake and you learn a lot more.
S = 1/2 a t^2
2.8 = 1/2 a (2.1)^2
2.8 = 2.205 a
a = 1.27 m/sec^2
m. a = F gravitational - F friction
m.a = m.g.sin 32.9 - m. g. cos 32.9 . kinetic coefficent
a = g.sin 32.9 - g. cos 32.9 . kinetic coefficent
1.27 = 10 x 0.54 - 10. 0.84 . kinetic coefficent
1.27 = 5.4 - 8.44 . kinetic coefficent
8.44 . kinetic coefficent = 5.4 - 1.27 = 4.13
kinetic coefficent = 4.13/8.44 = 0.489
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