A mass hangs from a fixed pulley. The mass is released from rest. Find the speed
ID: 1560090 • Letter: A
Question
A mass hangs from a fixed pulley. The mass is released from rest. Find the speed of the mass when it hits the ground. Assume the pulley is a solid disk or radius r = 0.25m. Two masses hang from a fixed pulley. The masses are released from rest with the heavier mass, m_2, starting at 1.7m above the floor. Find the speed of the masses when m_2 hits the ground. Assume the pulley is a hoop will all the mass on the outer edge (it has spokes of negligible mass). A cart, mass and pulley system are released from rest as shown. Find the velocity of the mass when it hits the ground.Explanation / Answer
total initial energy Ei = mb*g*h
total final enregy Ef = Ka + Kb
Ef = (1/2)*Ia*wa^2 + (1/2)*mb*vb^2
wa = vb/r
Ia = (1/2)*ma*r^2
Ef = (1/2)*(1/2)*ma*vb^2 + (1/2)*mb*vb^2
Ef = (ma/4 + mb/2)*vb^2
from energy conservation
total energy at any point is constant
Ef = Ei
(ma/4 + mb/2)*vb^2 = mb*g*h
(8/4 + 2.6/2)*vb^2 = 2.6*9.8*0.85
vb = 2.56 m/s <<<<<<------answer
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H2
total initial energy Ei = m2*g*h
total final enregy Ef = Kp + K2
Ef = (1/2)*Ip*wp^2 + (1/2)*m2*v2^2
wp = v2/r
Ip = mp*r^2
Ef = (1/2)*mp*v2^2 + (1/2)*m2*v2^2
Ef = (mp/2 + m2/2)*v2^2
from energy conservation
total energy at any point is constant
Ef = Ei
(mp/2 + m2/2)*v2^2 = m2*g*h
(15/2 + 11/2)*v2^2 = 11*9.8*1.7
v2 = 3.75 m/s <<<<<<------answer
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H3
total initial energy Ei = m2*g*h
total final energy Ef = K1 + K2 + Kp
Ef = (1/2)*m1*v2^2 + (1/2)*m2*v2^2 + (1/2)*Ip*wp^2
Ip = (1/2)*mp*r^2
wp = v2/r
Ef = (1/2)*m1*v2^2 + (1/2)*m2*v2^2 + (1/2)*(1/2)*mp*v2^2
Ef = (m1/2 + m2/2 + mp/4 )*v2^2
from energy conservation
total energy at any point is constant
Ef = Ei
(m1/2 + m2/2 + mp/4)*v2^2 = m2*g*h
(3.2/2 + 4.9/2 + 4.6/4)*v2^2 = 4.9*9.8*2.4
v2 = 4.71 m/s <<<<<<------answer
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