- #1

the_quack

- 9

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energy is conserved, no slipping or air resistance or anything.

There is a pulley, on one side is a mass resting on the ground, and on the other side is a larger mass above the ground. If the larger mass is released, what is it's speed when it hits.

okay, here is what I have:

PgM = Pgm + KM + KR + Km

Mgh = mgh + v[ srqt(.5M) + sqrt(.5I)/r + sqrt(.5m) ]

I=

.5mr^2 = .5(7.50kg)(.260m)^2 = 0.2535kg*m^2

h=3

m=18

M=26.5

so

PgM = 779.1 J

Pgm = 529.2 J

My answer is v(M)=49.84m/s, completely illogical, and the book answer is 3.22m/s. What did I do wrong? All the numbers look right, and it appears to make sense, but the answer is way off...

Thanks for any help!

There is a pulley, on one side is a mass resting on the ground, and on the other side is a larger mass above the ground. If the larger mass is released, what is it's speed when it hits.

okay, here is what I have:

PgM = Pgm + KM + KR + Km

Mgh = mgh + v[ srqt(.5M) + sqrt(.5I)/r + sqrt(.5m) ]

I=

.5mr^2 = .5(7.50kg)(.260m)^2 = 0.2535kg*m^2

h=3

m=18

M=26.5

so

PgM = 779.1 J

Pgm = 529.2 J

My answer is v(M)=49.84m/s, completely illogical, and the book answer is 3.22m/s. What did I do wrong? All the numbers look right, and it appears to make sense, but the answer is way off...

Thanks for any help!

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