A 2.1 kg block rests on a 30 slope and is attached by a string of negligible mas

Question

A 2.1 kg block rests on a 30 slope and is attached by a string of negligible mass to a solid drum of mass 0.80 kgand radius 6.9 cm , as shown in the figure (Figure 1) . When released, the block accelerates down the slope at 1.7 m/s2 .

What is the coefficient of friction between block and slope?

A 2.1 kg block rests on a 30 slope and is attached by a string of negligible mass to a solid drum of mass 0.80 kgand radius 6.9 cm , as shown in the figure (Figure 1) . When released, the block accelerates down the slope at 1.7 m/s2 .

What is the coefficient of friction between block and slope?

Explanation / Answer

The drum at the top of the slope and the string rotates the drum as the block slides.

2.1*g*(sin30 degree - cos30 degree * uk) - T = 2.1*1.7

Where, T is the string tension

since the string is on the rim of drum and doesn't slip

Summation(T) = I*alpha = 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

Summation(T) = T*r

T*r = 0.80*r^2*1.7/(2*r)

simplify

T = 0.80*1.7/2

T = 0.68 N

plug that in above

2.1*9.8*(sin30 degree - cos30 degree*uk) - 0.68 = 2.1*1.7

solve for uk

uk = (sin30 degree - (2.1*1.7+0.68)/(2.1*9.8))/cos30 degree

= 0.3389  

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