How Fast Does the Heavier Cart Move After Spring Release?

[marshall4]

A spring with a forces of 90N/m is compressed 8.8cm between two carts with mass of 1kg and 1.5kg. Friction is negligible, what is the final speed of the more massive of the bigger cart when the spring is released?

 

[PrudensOptimus]

v = +- v₀sqrt(1-x^2/A^2)

 

[M98Ranger]

To solve this problem, we can use the formula for potential energy stored in a spring: PE = 1/2kx^2, where k is the force constant and x is the displacement of the spring.

In this case, we are given the force constant (k = 90N/m) and the displacement (x = 8.8cm = 0.088m). Therefore, the potential energy stored in the spring is:

PE = 1/2 * 90N/m * (0.088m)^2 = 0.35 J

Since the system is assumed to have no energy losses due to friction, this potential energy will be converted into kinetic energy when the spring is released. We can use the formula for kinetic energy: KE = 1/2mv^2, where m is the mass of the cart and v is its final velocity.

Since we are interested in the final velocity of the more massive cart, we can set up the following equation:

0.35 J = 1/2 * (1.5kg) * v^2

Solving for v, we get v = 0.53 m/s. Therefore, the final speed of the more massive cart when the spring is released is 0.53 m/s.