- #1

SadDan

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## Homework Statement

Block 1 with mass m1=4.04 kg rests on a very low friction horizontal ledge. This block is attached to a string that passes over a pulley, and the other end of the string is attached to the hanging block 2 of mass m2=2.02 kg, as shown.

The pulley is a uniform disk of radius 11.85 cm and mass 1.980 kg. Calculate the speed of block 2 after it is released from rest and falls a distance of 1.84 m.

What is the angular speed of the pulley at the instant when block 2 has fallen a distance of 1.84 m ?

## Homework Equations

Wtot=change in Energy

KE=1/2 mv^2

W=integral of force*displacement

N2L for rotation and translation

## The Attempt at a Solution

The tension of 2 and 1 on the pulley should be different but the accelerations of the blocks would be the same?

a=m2*g/(mp/2 +m1+m2)

T2=m2(g-a)=13.27

T1=m1*a=13.0829

WEm2=integral from 0 to 1.84 (T2xdx) = 22.47 J

WT1m1=integral from 0 to 1.84 (T1xdx) = 22.1467 J

work energy theorem:

KE=W

1/2mv^2=WEm2+WT1m1

v=sqrt(2(WEm2+WT1m1)/mp)=3.33m/s

This is not the correct answer, I'm not sure what I am doing wrong would the energy side of the equation be only kinetic?