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4 years ago

6

We wrap a light, nonstretching cable around a 8.00 kg solid cylinder with diameter of 30.0 cm. The cylinder rotates with negligi

ble friction about a stationary horizontal axis. We tie the free end of the cable to a 13.0 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?

1 answer:

antoniya [11.8K]4 years ago

3 0

Answer:

h = 16.67m

Explanation:

If the kinetic energy of the cylinder is 510J:

$Kc=510=1/2*Ic*\omega c^2$

$\omega c=\sqrt{510*2/Ic}$

Where the inertia is given by:

$Ic=1/2*m_c*R_c^2=1/2*(8)*(0.15)^2=0.0225kg.m^2$

Replacing this value:

$\omega c=106.46rad/s$

Speed of the block will therefore be:

$V_b=\omega_c*R_c=106.46*0.15=15.969m/s$

By conservation of energy:

Eo = Ef

Eo = 0

$Ef = 510+1/2*m_b*V_b^2-m_b*g*h$

So,

$0 = 510+1/2*m_b*V_b^2-m_b*g*h$

Solving for h we get:

h=16.67m

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Since the projectile is going down the velocity at the instant it reaches the ground is:

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