Solving for the Speed of a Box on an Incline

AI Thread Summary
To solve for the speed of a 2.0 kg box on a frictionless incline at a 40-degree angle, the box is connected to a spring with a spring constant of 120 N/m. When the box moves 10 cm, it is assumed that the spring also compresses by the same distance. The energy conservation principle can be applied, where the potential energy lost by the box equals the spring's potential energy plus the kinetic energy of the box. The equation 1/2k(0.1)^2 = 1/2mv^2 - mg(10sin40) is suggested for calculating the speed. This approach effectively combines gravitational potential energy and spring potential energy to find the box's speed.
bodensee9
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Hello:

Can someone help with the following:
A 2.0 kg box on a frictionless incline of angle theta = 40 is connected by a cord that runs over a pulley to a light spring of spring constant k = 120N/m. The box is released from rest when the spring is unstretched. Assume pulley is massless and frictionless. I've attached drawing for clarification.

Find speed of box when it has moved 10 cm.

So can I assume that since the box has moved 10 cm, then the spring also has to move 10 cm? Then I can do
1/2k(.1)^2 = 1/2mv^2 - mg(10sin40).
Thanks.
 

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I can't the pic yet, but from the description, I think you're on the right track trying to do this with energy.
 

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