# Help finding mistake -

When mass M is at the position shown, it is sliding down the inclined part of a slide at a speed of 2.07 m/s. The mass stops a distance S2 = 1.85 m along the level part of the slide. The distance S1 = 1.23 m and the angle theta = 33.9 degrees. Calculate the coefficient (mu) of kinetic friction for the mass on the surface.

click on this link to view the picture

http://gryphon.phy.umassd.edu/msuphysicslib/Graphics/Gtype13/prob27a_MechEnWFriction.gif

What I did so far:

Since the mass stops at the end then
Work = E(final) - E(initial)
Work = KEf + PEf - KEi - PEi
Work = 0 + 0 - 0.5mv^2 - mgh -> h = S1sin(theta)
Work also = F.d -> in this case F is the friction and d is the total distance

now to find the mu -> coefficient of friction I separated the mass since F(friction) = mu.N
N being the normal force perpendicular to the surcace

when sliding on the slope: N = m.g.sin(theta)
So Friction Force = - mu.m.g.sin(theta)
Work done = W1 = - mu.m.g.sin(theta).S1

when sliding at the bottom N = m.g
So Friction Force = - mu.m.g
Work done = W2 = - mu.m.mg.S2

Total Work as calculated at first = Work = -0.5mv^2 - mg = W1 + W2

-mu.m.g.sin(theta).S1 - mu.m.mg.S2 = -0.5mv^2 - mgS1sin(theta)

mu = [ 0.5v^2 - gS1sin(theta) ] / [ gsin(theta)S1 + gS2 ] =

mu = 0.356

BUT IT IS WRONG !!
Help please.

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## Answers and Replies

when sliding on the slope: N = m.g.sin(theta)

Are you sure this is correct?

Delzac

oh ! No it's not correct ! N = m.g.cos(theta) !
Thanks for pointing that out !