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A simple atwood's machine uses 2 masses m1 and m2. Starting from rest, the speed of the two masses is 5.2m/s at the end of 6.7s. At that time, the kinetic energy of the system is 77J and each mass has moved a distance of 17.42 m.

The acceleration of gravity is 9.8m/s2.

a.)Find the value of the heavier mass
b.)Find the value of the lighter mass

2007-10-04 16:37:05 · 1 answers · asked by mathgeek 1 in Science & Mathematics Physics

1 answers

Oh that Atwood machine again.

What was that bloody equation of forces... Oh yea
The force moving the system equal to the difference in weights?
F=W1-W2
or
(m1+m2)a= m1g -m2g
(m1+m2)a= (m1 -m2)g

what is a? (I wonder)
a=(V-V0)/(t-t0) and the plot thickens with kinetic energy Ke
Ke=0.5(m1+m2)V^2

Ha I knew it! Two equation and two unknowns

(m1+m2)a= (m1 -m2)g or since a=V/t
(m1+m2)V/t= (m1 -m2)g rearanging it we get
(V/t -g)m1 + (V/t +g)m2=0

and Ke=0.5(m1+m2)V^2
0.5V^2 m1 + 0.5V^2 m2=Ke

can you solve for m1 and m2
(V/t -g)m1 + (V/t +g)m2=0
m1 + m2=2Ke/V^2

2007-10-05 06:58:44 · answer #1 · answered by Edward 7 · 0 0

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