Two Carts attached to a string

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B3NR4Y
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Homework Statement


Two 0.400-kg carts are 100 mm apart on a low-friction track. You push one of the carts with a constant force of 2.0N directed so that the cart you push moves away from the other cart.
Determine the acceleration of the center of mass of the two-cart system when the carts are connected by a spring of spring constant k = 300 N/m.

Homework Equations


[itex] F_{x} (x) = -k(x-x_{0}) [/itex] for the force of the spring
[itex] COM = \frac{\sum_n x_{n} m_{n}}{\sum_{n} m_{n}} [/itex] for the center of mass, initially is at 50 mm or 0.05 m
[itex] a_{cm} = \frac{\sum F(x)}{\sum m} [/itex] for the acceleration of the center of mass

The Attempt at a Solution


So the first part of the problem had me find the acceleration of the center of mass without the spring, which I did by summing the forces (which was easy, 2+0) and dividing by the sum of the masses, so without the spring the acceleration of the center of mass was 2.5 m/s/s. The second part of the problem I don't understand what it is attempting to say. If you push the right most cart 2 N, it will have a force acting in the other direction that is equal to -k(x-x0), but I don't know what the change in x is or anything like that so I am confused on where to begin.
 

Answers and Replies

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Orodruin
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And what is going to be the force on the other cart? What acceleration is it going to have?
 
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B3NR4Y
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The force on the other cart should be k(x-x0) because the spring is pulling it, if I am not wrong. Which is equal and opposite the the force the spring is exerting on the other cart, right?
 
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Orodruin
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Yes, so how are the accelerations of the carts changed by this? How does it affect the center of mass acceleration?
 
  • #5
B3NR4Y
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Yes, so how are the accelerations of the carts changed by this? How does it affect the center of mass acceleration?
I can't believe the solution was so simple, the spring part just cancels and you get the same exact answer as without the spring. Astounding, this is why I love physics. Thank you for your help, you're my favorite volcano :P
 
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Orodruin
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I can't believe the solution was so simple, the spring part just cancels and you get the same exact answer as without the spring.

Precisely. What matters for the acceleration of the CoM of a system is only its total mass and the external forces acting on it. In this case, the spring only provides an internal force and therefore need not be considered.

Astounding, this is why I love physics. Thank you for your help, you're my favorite volcano :p

You're welcome. Also, most people seem to blame me for helping in the creation of a certain piece of jewelry. They always conveniently forget that I was also instrumental in its destruction too! :p
 

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