Solving for the Speed of a Box on an Incline

[bodensee9]

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.

 

[G01]

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

 

[nlightner]

I would first clarify the given information and assumptions. It is important to note that the incline is frictionless and the pulley is both massless and frictionless. These conditions simplify the problem and allow us to use the equations of motion without accounting for any additional forces.

Next, I would address the question about the 10 cm movement of the box and the assumption that the spring also moves 10 cm. This assumption is correct as long as the cord connecting the box and the spring does not stretch or compress. In this case, the spring is light and the cord is assumed to be ideal, so this assumption is valid.

Now, to solve for the speed of the box, we can use the conservation of energy principle. The initial energy of the system consists of the gravitational potential energy of the box on the incline, which is converted into the kinetic energy of the box as it moves down the incline and the potential energy of the compressed spring. Therefore, we can set the initial energy equal to the final energy and solve for the speed of the box.

Using the equation provided in the question, we can set the initial energy equal to 1/2k(.1)^2, since the spring is initially unstretched. The final energy can be written as 1/2mv^2 + 1/2kx^2, where m is the mass of the box, v is the speed of the box, k is the spring constant, and x is the distance the spring has been compressed (in this case, 10 cm). We can then solve for v by setting the initial energy equal to the final energy and solving for v.

I would also suggest checking the units in the equation to ensure they are consistent. In this case, the units for the spring constant should be N/m and the units for the distance should be meters. Double checking the units can help catch any errors in the calculations.

In conclusion, to find the speed of the box on the incline after it has moved 10 cm, we can use the conservation of energy principle and solve for the speed using the equation 1/2k(.1)^2 = 1/2mv^2 + 1/2kx^2. This method can be applied to similar problems with different variables or conditions.