Solving for the Speed of a Box on an Incline

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The discussion focuses on calculating the speed of a 2.0 kg box on a frictionless incline at an angle of 40 degrees, which is connected to a spring with a spring constant of 120 N/m. The box is released from rest and moves 10 cm down the incline. The user correctly identifies that the energy conservation principle can be applied, equating the potential energy lost by the box to the elastic potential energy stored in the spring and the gravitational potential energy. The formula used is 1/2k(0.1)^2 = 1/2mv^2 - mg(10sin40).

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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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