Can 600 N Pull the Same Cart Twice? - Get a Second Opinion!

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The discussion centers on comparing the acceleration of a cart being pulled by a man exerting a force of 600 N versus an iron weight of 60 kg attached to a rope. Both scenarios result in the same force of 600 N acting on the cart, leading to the conclusion that the acceleration will be identical in both cases. The calculations confirm this, as the acceleration is derived using the formula a = F/m, yielding a consistent value of 12 m/s². The angle of tension does not affect the acceleration as long as it remains constant. Overall, the analysis supports that both methods produce the same acceleration for the cart.
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Force Question - Still want another oppinion!

A man pulls a cart on a friction fee surface by means of a pulley and rope. He pulls the rope at 600 N. Now replace the man's action with an iron weight of 60 kg so that it is attached to the end of the rope. The cart is again set in motion. Is the acceleration the same in both situations. ( Use g to 10m/s/s) , (mass of cart = 50kg)

You can see a picture of the situation here: http://www.members.shaw.ca/qwex/physics.jpg

So my assumption is that both the man and the weight would pull with 600 N downwards. So would this make the acceleration the same. I'm guessing the distance is the same in both situations as well. Is there any way I can explain this? Am I even correct? Thank you!
 
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I think you are right, the acceleration would be the same since to determine the force that the iron pulls the cart. We use the formula:
F = ma = 60 x 10 = 600 N which is the same as the force that the man pulls the rope.

XMLT
 
Thanks! Any one else?
 
Acceleration will be the same, it depends only on the force applied to it. If the angle, at which tension is prodced, is same then the acc. will be the same.
To show it correct prove it mathematically or go to the experiments.
a = f/m = 600/50 = 12 (man pulled)
a = f/m = (M*g)/m = 60*10/50 = 12
 
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