Please show step and make answer clear. (Box the answer with corresponding numbe

Question

Please show step and make answer clear. (Box the answer with corresponding number. ) Thank you.

5. EXTRA CREDIT. (11 POINTS) A 1.000-kg block hangs vertically at the end of a string wrapped around a pulley of radius R= 0.250 m and mass M-2.000-kg shaped in the form of a solid cylinder. Thus, the pulley has I MR2 about the axis of rotation through the center. The vertically hanging block and pulley are shown at the start of the motion when they are released from rest. The block is released from a height h = 1.000 m above the ground. What is the linear speed v of the block just before it hits the ground? 1 kg

Explanation / Answer

using law of law of conservation of energy

energy of the system when they are at h = 1 m = energy of the system when they were hitting the ground

(m*g*h) = (0.5*m*v^2)+(0.5*I*w^2)

I = 0.5*M*R^2

and angular speed is w = V/R

then

m*g*h =(0.5*m*V^2)+(0.5*0.5*M*R^2*(V/R)^2)

(1*9.8*1) = (0.5*1*V^2)+(0.5*0.5*2*V^2)

9.8 = 0.5*V^2 + 0.5*V^2

9.8 = V^2

v = sqrt(9.8) = 3.13 m/sec

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---