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**Calculating spring equilibrium length HW question.... HARD!!!!**

## Homework Statement

A spring is connected to the bottom of a ramp inclined at an angle of 45 degrees. In its equilibrium position, the spring is equal to the length of the ramp, L. If the spring constant is 500 N/m and I want to launch a 5 kg ball, 100 meters as measured from the base of the ramp, how long must the ramp be if my initial compression of the spring is equal to 1/2 the length of the ramp. AND. friction on the ramp cannot be ignored.... the coefficient of friction is .25

## Homework Equations

I know that the spring potential equals 1/2kx^2 and the Work due to friction will equal mgcos45*L. The Total Energy at the top of the ramp will be equal to the Energy from the Spring - Work done by friction

## The Attempt at a Solution

I think I understand the concepts of the problem, but I just don't know how to putit all together... You can find the Spring potential and subtract the Work done by friction to give you an energy at the top of the ramp... This energy can be converted to Kinetic energy of the ball by equaling it to 1/2mv^2. Then you can solve for a velocity. This velocity will have an x and y component which then can be used to determine the time in the y direction until it hits the ground and then this time can be used to find the range in the x direction. BUT AGAIN I can't get it all put together.