Question

ip a 1.2 kg block is tied to a string that is wrapped around the rim of a pulley of radius 7.1cm. the block is released from rest. EXPLAIN

l, LhandlerSig Hide Timer Part A Assuming the pulley is a uniform disk with a mass of 0.31 kg, find the speed of th block after it has fallen through a height of 0.50 m Express your answer using two significant figures. 2 m/s Submit Submit Request Answer Part B 1033 AM di ^

Answer 1

here,

mass tied , m1 = 1.2 kg

radius of pulley , r = 7.1 cm = 0.071 m

A)

mass of pulley , m2 = 0.31 kg

the acceleration of system , a = net force /effective mass

a = ( m1 * g) /( m1 + (0.5 * m2 * r^2)r^2)

a = ( 1.2 * 9.81) / ( 1.2 + (0.5 * 0.31))

a = 8.69 m/s^2

h = 0.5 m

the speed of block after it is fallen 0.5 m , v = sqrt(2 * a * h)

v = sqrt(2 * 8.69 * 0.5)

v = 2.95 m/s

B)

when a mass is attached to the rim of the pulley

the effective mass of the system increases

so, the acceleration of system decreases

as v = sqrt( 2 * a * h)

when acceleration decreases , the speed of block DECREASES