MasteringPhysics Chapter 12-Google Chrome https// ignmentProblemiD-56100031 sess

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MasteringPhysics Chapter 12-Google Chrome https// ignmentProblemiD-56100031 session.masteringphysics.com/myct/itemView?ass Phys201-2016-Spring Chapte 12 Problem 12 69 s previous 9 of 15 | net Problem 12 69 Part A t the blocks are relesed from rest, how long does it take the 40 kg block to reach the floor? The two blocks in the figurelige.1) are connected by a massless rope that passes over a pulley. The pulley is 17 cn in diamoter and has a mass of 2 5 kg As the pulley tuns, friction at the axle exerts a torque of magnhude 0 48 N·m Express your answer to two significant figures and include the appropriate units I Value Units Submit Continue 4.0kg 1.0 m 20 kg 447 P 4/10/20

Explanation / Answer

mass at the left hand,m1= 4.0 kg

Tension at the left hand,T1

mass at the right hand,m2= 2.0 kg

mass of the pulley,m3 = 2.5 kg

Right hand tension is T2

g (be the acceleration due to gravity) = 9.78 m/s2

acceleration of masses is a

the angular acceleration of the pulley is alpha

the radius of the pulley, r = 0.17 m

the friction ,F = 0.48 N.m

the hight for m1, h = 1 m

the time taken to reach the floor is t

For mass m1:

m1 g - T1 = m1 a

T1 = m1(g - a)

For mass m2:

T2 - m2 g = m2 a

T2 = m2(g + a)

For the pulley:

(T1 - T2)r - F = I alpha

= (m3 r2 / 2)(a / r)

T1 - T2 = F / r + m3 a / 2

*Eliminate T1 and T2:

m1(g - a) - m2(g + a) = F / r + m3 a / 2

(m1 - m2)g - (m1 + m2)a - m3 a / 2 - F / r

(m1 + m2 + m3 / 2)a = F / r + (m1 - m2)g

a = [ F / r + (m1 - m2)g ] / (m1 + m2 + m3 / 2)

h = at2 / 2

t = sqrt(2h / a)

= sqrt{ h(2m1 + 2m2 + m3) / [ F / r + (m1 - m2)g ] }

so t = sqrt{ 1( 2 * 4 + 2 * 2 + 2.5) / [ 0.48 / 0.17 + (4 - 2)9.78 ] }

= 0.804 sec

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