A mass m_1 = 5.6 kg rests on a frictionless table. It is connected by a massless

Q: A mass m_1 = 5.6 kg rests on a frictionless table. It is connected by a massless

for m2 T - m2*g = m2*a T2 = m2*g + m2*a for m1 m1*g*sin70 - T = m1*a m1*g*sin70 - m2*g - m2*a = m1*a a = (m1*sin70 - m2)*g/(m1+m2) a = ((5.6*sin70) - 3.8)*9.8/(5.6+3.8) a = 1.52 m/s^2

ID: 1434773 • Letter: A

Question

A mass m_1 = 5.6 kg rests on a frictionless table. It is connected by a massless and frictionless pulley to a second mass m_2 = 3.8 kg that hangs freely. What is the magnitude of the acceleration of block 1? m/s^2 What is the tension in the string? Now the table is tilted at an angle of theta = 69degree with respect to the vertical. Find the magnitude of the new acceleration of block 1. At what "critical" angle will the blocks NOT accelerate at all? Now the angle is decreased past the "critical" angle so the system accelerates in the opposite direction. If theta = 20degree find the magnitude of the acceleration. m/s^2

Explanation / Answer

for m2

T - m2*g = m2*a

T2 = m2*g + m2*a

for m1

m1*g*sin70 - T = m1*a

m1*g*sin70 - m2*g - m2*a = m1*a

a = (m1*sin70 - m2)*g/(m1+m2)

a = ((5.6*sin70) - 3.8)*9.8/(5.6+3.8)

a = 1.52 m/s^2

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A mass m_1 = 5.6 kg rests on a frictionless table. It is connected by a massless

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