A 5.00 kg block sits on an incline of 36.9 with a coefficient of kinetic frictio

Question

A 5.00 kg block sits on an incline of 36.9 with a coefficient of kinetic friction k = 0.25. Attached to the block is a string, which is wound around a massive frictionless pulley (also called a flywheel) at the top of the incline. The flywheel has a mass of 25.0 kg and a moment of inertia of 0.500 kg·m2. a) What is the radius of the flywheel? b) Draw a free body diagram (or extended free body diagram, if appropriate) for each of the two objects. c) What is the acceleration of the block down the incline? d) What is the tension in the string?

Explanation / Answer

Given values are,

Mass of block, m = 5 kg

= 36.9o

Coefficient of kinetic friction, k = 0.25

Mass of flywheel, M = 25 kg

Moment of inertia, I = 0.05 kg.m2

a)

Moment of inertia is,

I = Mr2 where r is the radius of flywheel

So r2 = I/M = 0.05 / 25 = 0.002

So, r = 0.044 m = 4.4 Cm

Therefore,

Radius of flywheel is 4.4 cm.

b)

Refer this diagram,

c) For acceleration,

We must know about downward (slope) force.

By component form,

F = mg sin

But F = ma,

So ma = mg sin

Therefore,

Acceleration a = g sin

= 9.8 * sin(36.9) = 9.8 * 0.6

= 5.88 m/s2

d) Tension in the string is,

T - mgsin = ma

So, T = ma + mgsin

= 2 ma

= 2 * 5 * 5.88

= 58.8 N (Where a = gsin )

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.