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aten_vs_ra
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Relevant equations: Newton’s 2nd and 3rd Laws. Hooke’s Law.
A 4kg cart and a 6kg cart are connected by a relaxed horizontal spring with spring constant 100 N/m. You pull the 6kg cart with some constant horizontal force. After a time, the separation between the carts remains constant and the spring is stretched by 0.1m.
(a) Carefully (with forces drawn to the same scale on both diagrams) draw free body diagrams for after the spring has reached constant stretch for each of the two carts, labeling all relevant forces and showing the direction of the acceleration.
The freebody diagrams I’ve drawn do not feature gravity or the normal force of the carts on the ground as I don’t think they’re hugely relevant to the given problem. (The posted answers to our previous homework set were confined to one axis.)
Both carts will have the same acceleration, direction and magnitude. Let’s say 6kg cart is on the right and make that our positive direction. I think this follows from the fact that the separation between the carts remains constant after a while.
The forces acting on the 4kg cart will be the friction of the ground (pointing left) and the tension force of the spring (point right). The tension force must be greater than the friction force exerted by the ground for the cart to accelerate right.
The forces acting on the 6kg cart will be the friction force of the ground (pointing left), the tension force of the spring (same magnitude as the tension force operating on the 4kg cart, but opposite in direction now, pointing left), and the pulling force you exert on the cart (pointing right). The pulling force must be greater than the sum of the friction force and the tension force if the 6kg cart is to be accelerating.
I’m an unsure of the relative sizes of the friction forces. We haven’t gotten into them yet, but it’s the product of the normal force and a coefficient of friction? If so, then the 6kg cart is going to be experiencing a greater friction force then the 4kg, so I should be sure to draw the arrows appropriately.
(b) What pulling force did you apply to the 6kg cart to reach this constant stretch?
I know I can determine the tension force of the spring from Hooke’s law (think its 10 N), and that it’ll be positive for 4kg cart but negative for 6kg cart. I think the idea is that I’m supposed to take this information and use Newton’s second law to find out the acceleration of 4kg cart (it only has two forces so it should be simpler to work with) and use this information for the 6kg cart, since it should have the same acceleration. I don’t quite know how the friction forces on the carts fit into this. As I said earlier, we haven’t gone into great detail about them. I’m kind of flummoxed as to how to set this up. It seems like I have too many unknowns.
Thank you for any consideration.
A 4kg cart and a 6kg cart are connected by a relaxed horizontal spring with spring constant 100 N/m. You pull the 6kg cart with some constant horizontal force. After a time, the separation between the carts remains constant and the spring is stretched by 0.1m.
(a) Carefully (with forces drawn to the same scale on both diagrams) draw free body diagrams for after the spring has reached constant stretch for each of the two carts, labeling all relevant forces and showing the direction of the acceleration.
The freebody diagrams I’ve drawn do not feature gravity or the normal force of the carts on the ground as I don’t think they’re hugely relevant to the given problem. (The posted answers to our previous homework set were confined to one axis.)
Both carts will have the same acceleration, direction and magnitude. Let’s say 6kg cart is on the right and make that our positive direction. I think this follows from the fact that the separation between the carts remains constant after a while.
The forces acting on the 4kg cart will be the friction of the ground (pointing left) and the tension force of the spring (point right). The tension force must be greater than the friction force exerted by the ground for the cart to accelerate right.
The forces acting on the 6kg cart will be the friction force of the ground (pointing left), the tension force of the spring (same magnitude as the tension force operating on the 4kg cart, but opposite in direction now, pointing left), and the pulling force you exert on the cart (pointing right). The pulling force must be greater than the sum of the friction force and the tension force if the 6kg cart is to be accelerating.
I’m an unsure of the relative sizes of the friction forces. We haven’t gotten into them yet, but it’s the product of the normal force and a coefficient of friction? If so, then the 6kg cart is going to be experiencing a greater friction force then the 4kg, so I should be sure to draw the arrows appropriately.
(b) What pulling force did you apply to the 6kg cart to reach this constant stretch?
I know I can determine the tension force of the spring from Hooke’s law (think its 10 N), and that it’ll be positive for 4kg cart but negative for 6kg cart. I think the idea is that I’m supposed to take this information and use Newton’s second law to find out the acceleration of 4kg cart (it only has two forces so it should be simpler to work with) and use this information for the 6kg cart, since it should have the same acceleration. I don’t quite know how the friction forces on the carts fit into this. As I said earlier, we haven’t gone into great detail about them. I’m kind of flummoxed as to how to set this up. It seems like I have too many unknowns.
Thank you for any consideration.
