The two blocks in (Figure 1) are connected by a massless rope that passes over a
ID: 1438969 • Letter: T
Question
The two blocks in (Figure 1) are connected by a massless rope that passes over a pulley. The pulley is 14 cm indiameter and has a mass of 1.8 kg . As the pulley turns, friction at the axle exerts a torque of magnitude 0.40 N?m.
If the blocks are released from rest, how long does it take the 4.0 kg block to reach the floor?
The two blocks in (Figure 1) are connected by a massless rope that passes over a pulley. The pulley is 14 cm indiameter and has a mass of 1.8 kg . As the pulley turns, friction at the axle exerts a torque of magnitude 0.40 N?m.
If the blocks are released from rest, how long does it take the 4.0 kg block to reach the floor?
MasteringPhysics: Ch 07 HW-Google Chrome https://session.masteringphysics.com/myct/itemView?assignmentProblemID=58 1861 27&offset;=next General Physics I - Spring 2016 Signed in as bayan al yaqoob Help Close Ch 07 HWProblem 7.69 Resources previous | 8 of 9 | next » Problem 7.69 Part A The two blocks in (Figure 1) are connected by a massless rope that passes over a pulley. The pulley is 14 cm in diameter and has a mass of 1.8 kg As the pulley turns friction at the axle exerts a torque of magnitude 0.40 N·m If the blocks are released from rest, how long does it take the 4.0 kg block to reach the floor? Express your answer using two significant figures. Figure 1 of 1 Submit My Answers Give Up vide Feedback Continue 4.0 kg 1.0 m 2.0 kg Search the web and Windows 0 3/17/2016Explanation / Answer
Sum forces about each block and torques about the pulley to find the acceleration
Then use kinematic eqn to find time
So m1*g - T1 = m1*a or T1 = m1*g - m1*a
for the 2kg block we have T2 - m2*g = m2*a or T2 = m2*g + m2*a
for the pulley we have T1*r - 0.40 - T2*f = I* where I = 1/2*M*r^2 and = a/r
so we have (m1*g - m1*a)*r - 0.40 -(m2*g + m2*a) *r = 1/2*M*r^2*a/r = 1/2*M*r*a
rearranging terms
a*(1/2*M*r + m1*r + m2*r) = m1*g*r - m2*g*r -0.70
or a*(1/2*1.8*0.07 + 4.0*0.07 + 2.0*0.07) = 4*9.8*0.07 - 2.0*9.8*0.07 - 0.40
so a = 0.972/0.483 = 2.012m/s^2
So the 4 kg block must fall 1.0m from rest under an acceleration of 2.012m/s^2
y = 1/2*a*t^2
or t = sqr(2*y/a) = sqrt(2*1.0/2.012) = 0.99s
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