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Ressages Courses Hep Logout Hain course contents »...-week 9 Homework set-due Friday Acceleration in an Atwood's Machine In an Atwoods machine, one block has a mass of 782.0 g, and the other a mass of 972.0 g. The pulley, which is mounted in horizontal frictionless bearings, has a radius of 5.70 cm. when released frem rest, the heavier block is observed to fall 61.0 cm in 1.93 s (without the string slipping on the pulley). MI what is the magnitude of the acceleration of the 782.0-g block? What is the magnitude of the acceleration of the 972.0-g block? Submt Answer Tries 0/10 Subm Answer Tries /10 iets What is the magnitude of the tension in the part of the cord that supports the 782.0-g block? Submit Answer Tries 0/10 what is the magnitude of the tension in the part of the cord that supports the 972.0-g block? Submit Answer Tries Q'10 what is the magnitude of the angular acceleration of the pulley? Submt Answer Tries 0/10 What is the rotational inertia of the pulley? Submit Answer Tries 0/10 What is the change in the potential energy of the system after 1.93 s? Submit Answer Tries 0/10 Send Feedbeck Püst Discussion oMVw

Explanation / Answer

here,

m = 0.782 kg

M = 0.972 kg

radius , r = 0.057 m

h = 0.61 m

time taken , t = 1.93 s

a)

let the accelration be a

h = 0 + 0.5 * a * t^2

0.61 = 0 + 0.5 * a * 1.93^2

a = 0.33 m/s^2

the magnitude of accelration of 782 g is 0.33 m/s^2

b)

the magnitude of accelration of 972 g is 0.33 m/s^2

c)

for m

T = m * ( g + a)

T = 0.782 * ( 0.33 + 9.81) N

T = 7.92 N

d)

the magnitude of tension is 7.92 N

e)

the angular accelration of pulley , alpha = a/r

alpha = 5.79 rad/s

f)

let the rotational inertia be I

accelration , a = (M - m) * g /( m + M + I/r^2)

0.33 = ( 0.972 - 0.782) * 9.81 /( 0.972 + 0.782 + I/0.057^2)

I = 1.27 * 10^-2 kg.m^2

g)

the potential energy lost , PE = g * h ( M - m)

PE = 9.81 * 0.62 * ( 0.972 - 0.782) J

PE = 1.16 J

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