An 8.0 kg mass, M and a 2kg mass m are suspended vertically from the top of a massless, frictionless pulley using massless rope. Assume the system is initially at rest and at t=0 mass, M is released (in the diagram they are both hanging straight down with mass, m hanging lower than mass, M).
a) use Newton's second law of the motion of mass to write an equation for M.
I got F = 8a, is that right?
b) Repeat part a for the motion of mass m
I got F = 2a
c) Solve for the magnitude of the acceleration of each block.
I have no idea how to do this!! I was trying to cancel out the Fnet.. but I think I over thought it..I don't know.
d) If M drops 1.5m how much work is done by gravity on the two block system.
Can someone help me???
a) use Newton's second law of the motion of mass to write an equation for M.
I got F = 8a, is that right?
b) Repeat part a for the motion of mass m
I got F = 2a
c) Solve for the magnitude of the acceleration of each block.
I have no idea how to do this!! I was trying to cancel out the Fnet.. but I think I over thought it..I don't know.
d) If M drops 1.5m how much work is done by gravity on the two block system.
Can someone help me???
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You forgot about gravity.
I can't do this in my head quickly, but basically there is a downwards force 8*g on M and a downwards force 2*g on m, then an upwards forcein the rope which must be the same for both, and an acceleration of both that must be equal and opposite
So I think it's F = 8(g-a) = 2 (g+a)
I can't do this in my head quickly, but basically there is a downwards force 8*g on M and a downwards force 2*g on m, then an upwards forcein the rope which must be the same for both, and an acceleration of both that must be equal and opposite
So I think it's F = 8(g-a) = 2 (g+a)