physics exam - Page 2

quote:

ORIGINAL: Nauree

Ok this is a beast homework problem.

A crate of mass m1 on a frictionless inclined plane is attached to another crate of mass m2 by a massless rope. The rope passes over an ideal pulley so the mass m2 is suspended in air. The plane is inclined at an angle = 36.9°. Use conservation of energy to find how fast crate m2 is moving after m1 has traveled a distance of 1.4 m along the incline, starting from rest. The mass of m1 is 10.2 kg and the mass of m2 is 18.4 kg.

I'm trying to figure out what types of energy each have. I believe m2 only have gravitional potential energy but I'm unsure of what if m1 has kinetic energy. I know they both have to have grav potential. Right now I've concluded that that the Work for m2 = Work for m1.

I'm sitting here in the dark, infront of my computer, with my cal, playing rock and metal. Extreme Physics!

 
You're making it way too hard. It's a basic conversion from potential energy to kinetic energy.
M2 moves 1.4m closer to the ground, M1 moves 1.4m*sin(36.9°) away from the ground.
Change in potential energy= m*g*delta h
m2 change = 18.4kg * 9.81 m/s/s * 1.4m = 252.7J
m1 change = 10.2kg * 9.81 m/s/s * -0.841m = -84.2J
 
Now that energy (252.7J - 84.2J) has been converted from potential to kinetic energy. Kinetic energy is 1/2 m*v^2 so...
252.7-84.2 = 168.6 (total energy that is now kinetic)
168.6J=.5*(18.4+10.2)*v^2 (the two masses are tied together with the massless rope, they act as one object because one of them can't be moving faster then the other)
v^2=11.8
v=3.4 m/s
 
To prove this to yourself you can do it the old fashioned way, find the acceleration of m2, find the time it would take for it to move 1.4m, find speed at 1.4m. This is basically just an exercise in showing you that conservation of energy problems can save you a lot of work from doing a problem a different way.

quote:

ORIGINAL: therabbit

quote:

ORIGINAL: Nauree

Ok this is a beast homework problem.

A crate of mass m1 on a frictionless inclined plane is attached to another crate of mass m2 by a massless rope. The rope passes over an ideal pulley so the mass m2 is suspended in air. The plane is inclined at an angle = 36.9°. Use conservation of energy to find how fast crate m2 is moving after m1 has traveled a distance of 1.4 m along the incline, starting from rest. The mass of m1 is 10.2 kg and the mass of m2 is 18.4 kg.

I'm trying to figure out what types of energy each have. I believe m2 only have gravitional potential energy but I'm unsure of what if m1 has kinetic energy. I know they both have to have grav potential. Right now I've concluded that that the Work for m2 = Work for m1.

I'm sitting here in the dark, infront of my computer, with my cal, playing rock and metal. Extreme Physics!

You're making it way too hard. It's a basic conversion from potential energy to kinetic energy.
M2 moves 1.4m closer to the ground, M1 moves 1.4m*sin(36.9°) away from the ground.
Change in potential energy= m*g*delta h
m2 change = 18.4kg * 9.81 m/s/s * 1.4m = 252.7J
m1 change = 10.2kg * 9.81 m/s/s * -0.841m = -84.2J

Now that energy (252.7J - 84.2J) has been converted from potential to kinetic energy. Kinetic energy is 1/2 m*v^2 so...
252.7-84.2 = 168.6 (total energy that is now kinetic)
168.6J=.5*(18.4+10.2)*v^2 (the two masses are tied together with the massless rope, they act as one object because one of them can't be moving faster then the other)
v^2=11.8
v=3.4 m/s

To prove this to yourself you can do it the old fashioned way, find the acceleration of m2, find the time it would take for it to move 1.4m, find speed at 1.4m. This is basically just an exercise in showing you that conservation of energy problems can save you a lot of work from doing a problem a different way.

What would the entire conservation equal be? Our prof is making is put every type of energy associated even if they come out to 0.

Like would

E1 = E2
U1 + K1 = U1 + K2
How would each equation cancle out to only be left with just Kinetic energy.

dear god we got a lot of nerds on this forum. haha

im probably one of them too!!

quote:

ORIGINAL: Nauree
What would the entire conservation equal be? Our prof is making is put every type of energy associated even if they come out to 0.

Like would

E1 = E2
U1 + K1 = U1 + K2
How would each equation cancle out to only be left with just Kinetic energy.

Total conservation would be
(U1,1) + (U1,2) + (K1,1) + (K1,2) = (U2,1) + (U2,2) + (K2,1) + (K2,2)
Total energy at the start equals the total energy at the end
K1,1 and K1,2 are both 0 because at the start everything is stationary, and you don't know what any of the potential energies are, you only know the differences.
So rewrite the equation like this:
[(U1,1)-(U2,1)] + [(U1,2)-(U2,2)] = [(K2,1)+(K2,2)]
[18.4kg * 9.81 m/s/s * 1.4m] + [10.2kg * 9.81 m/s/s * -0.841m] = [0.5*(18.4+10.2)*3.4^2]
[252] + [-84] = [168]
[168]=[168]
Thus, energy is conserved. Tada!