A 3.0 k g block rests on a 30 ? slope and is attached by a string of negligible

Question

A 3.0kg block rests on a 30 ? slope and is attached by a string of negligible mass to a solid drum of mass 0.50kg and radius 6.3cm , as shown in the figure (Figure 1) . When released, the block accelerates down the slope at 1.6m/s2 .

What is the coefficient of friction between block and slope?

A 3.0kg block rests on a 30 ? slope and is attached by a string of negligible mass to a solid drum of mass 0.50kg and radius 6.3cm , as shown in the figure (Figure 1) . When released, the block accelerates down the slope at 1.6m/s2.

Explanation / Answer

For the block, the sum of forces...

Weight component acting down the hill = mgsin(angle)

Tension acting up the hill = T

Friction acting up the hill = umgcos(angle)

The tension provides a torque such that Tr = I(alpha)

I for the disk is .5mr^2 and alpha = a/r, so

Tr = .5Mr^2a/r

T = .5Ma

Now...

mgsin(30) - .5Ma - umgcos(30) = ma

(3)(9.8)(sin 30) - .5(.5)(1.6) - (u)(3)(9.8)(cos 30) = (3)(1.6)

u = .373

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