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## Homework Statement

A simple Atwood's machine uses two masses m1 and m2. Starting from rest, the speed of the two masses is 8.7 m/s at the end of 7.4 s. At that time, the kinetic energy of the system is 61 J and each mass has moved a distance of 32.19 m.

Find the value of the heavier mass.

Find the value of the lighter mass.

## Homework Equations

a = (m

_{2}- m

_{1}) / (m

_{1}+ m

_{2})g

Ke = 1/2mv

^{2}

## The Attempt at a Solution

61 = 1/2m

_{1}v

^{2}+ 1/2m

_{2}v

^{2}

61 = 1/2(8.7)

^{2}(m

_{1}+m

_{2})

61 = 37.845(m

_{1}+m

_{2})

m

_{1}+m

_{2}= 1.612

a = 8.7 / 7.4 = 1.176

1.176 = (m

_{2}- m

_{1}) / (m

_{1}+ m

_{2})g

10.976m

_{1}= 8.624m

_{2}

m

_{1}= .786m

_{2}

.786m

_{2}+ m

_{2}= 1.612

2m

_{2}= 2.051

m

_{2}= 1.025

m

_{1}+ 1.025 = 1.612

m

_{1}= .587

I was sure that I solved it correctly, perhaps my algebra for solving the acceleration equation is incorrect.

Thanks for any help.