hmmm I still don't get the answer.. I think the trouble comes in because maybe I am not combining the forces in the x direction and forces in the y direction correctly?
In the y-direction, you have m
1g acting downward on the object and T acting upward on the object. The net force on the downward object is m
1g - T.
In the x-drection, you have the man accelerating across the frictionless surface, being impacted solely by T. His normal force is offset by his weight, so we need only consider his x-direction forces. Because the surface is frictionless, the only x-direction force acting on the man is tension.
Before we start the math, it's important to note that we have a
man = a
object in magnitude, because they are connected by the rope.
In y: m
1a = m
1g - T
In x: m
2a = T
Plugging m
2a for T in the y-direction equation gives:
m
1a = m
1g - m
2a
So,
m
1a + m
2a = m
1g
(m
1 + m
2)a = m
1g
a = m
1/(m
1 + m
2) x g
a = [50/(100 + 50)] x g = 1/3 x g = 3.3
Plugging into the x-direction relationship gives:
T = m
2a = m
2(1/3g)
T = 100 x 3.3 = 333 N
This is an awful lot of math for one problem, so once you feel semi-comfortable with this answer, do the question again by POE. You never said what the answer choices are, but I'd be willing to bet you can eliminate two choices for sure if not three.
The object is falling, so T must be less than 500N (otherwise T would cancel the weight of the object and it wouldn't fall). T must have some value, otherwise the man and object would be accelerating at g.
