Question

# A 3.0 kg block rests on a 30 ∘ slope and is attached by a string...

A 3.0 kg block rests on a 30 ∘ slope and is attached by a string of negligible mass to a solid drum of mass 0.80 kg and radius 7.5 cm , as shown in the figure (Figure 1) . When released, the block accelerates down the slope at 2.0 m/s2 . What is the coefficient of friction between block and slope?

I got .36 every time I worked it out and it kept saying it was wrong. If someone can get a different answer please help me out.

according to given figure

3.0*g*(sin30-cos30*µk)-T=3.0*2.0
Where T is the string tension
since the string is on the rim of drum and doesn't slip

Στ=I*α=I*a/r
where I is the moment of inertia of the drum and a is the acceleration of the block
For the drum
I=0.80*r^2/2
Στ=T*r
T*r=0.80*r^2*2.0/(2*r)
simplify

T=0.8 N

plug that in above

3.0*g*(sin30-cos30*µk)-0.8=3.0*2.0
=> 29.4sin30 - 29.4cos30*µk = 6.8

=> 7.9 = 29.4cos30*µk

=> 7.9 = 25.46*µk

=> µk = 0.3102

µk=(sin30-(2.8*1.3+0.39)/
(2.8*9.8))/cos30

µk=0.407764

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