We wrap a light, nonstretching cable around a 10.0 kg kg solid cylinder with diameter of 38.0 cm cm. The cylinder rotates with negligible friction about a stationary horizontal axis. We tie the free end of the cable to a 14.0 kg kg block and release the block from rest. As the block falls, the cable unwinds without stretching or slipping. How far will the mass have to descend to give the cylinder 510 J of kinetic energy?

Question

Ariel Odonnell

Physics

24 July, 06:56

We wrap a light, nonstretching cable around a 10.0 kg kg solid cylinder with diameter of 38.0 cm cm. The cylinder rotates with negligible friction about a stationary horizontal axis. We tie the free end of the cable to a 14.0 kg kg block and release the block from rest. As the block falls, the cable unwinds without stretching or slipping. How far will the mass have to descend to give the cylinder 510 J of kinetic energy?

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Answers (1)

Know the Answer?

Edith Carney · Answer 1

Answer: 14.16

Explanation:

Given

d = 38cm

r = d/2 = 38/2 = 19cm = 0.19m

K. E = 510J

m = 10kg

I = 1/2mr²

I = 1/2*10*0.19²

I = 0.18kgm²

When it has 510J of Kinetic Energy then,

510J = 1/2Iω²

ω² = 1020/I

ω² = 1020/0.18

ω² = 5666.67

ω = √5666.67 = 75.28 rad/s

Velocity is the block, v = ωr

V = 75.28 * 0.19

V = 14.30m/s

The "effective mass" M of the system is

M = (14.0 + ½*10.0) kg = 19.0 kg

The motive force would be

F = ma

F = 14 * 9.8

F = 137.2N

so that the acceleration would be

a = F/m

a = 137.2/19

a = 7.22m/s²

Finally, using equation of motion.

V² = u² + 2as

14.3² = 0 + 2*7.22*s

204.49 = 14.44s

s = 204.49/14.44

s = 14.16m