- #1

B3NR4Y

Gold Member

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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-x

_{0}), but I don't know what the change in x is or anything like that so I am confused on where to begin.