# Calculating a force constant using kinetic and potential energy?

1. Oct 3, 2005

### erik-the-red

You are designing a delivery ramp for crates containing exercise equipment. The crates weighing $$1500 N$$ will move at a speed of $$2.00 m/s$$ at the top of a ramp that slopes downward at an angle $$24.0^\circ.$$ The ramp exerts a kinetic friction force of $$540 N$$on each crate, and the maximum static friction force also has this value. Each crate will compress a spring at the bottom of the ramp and will come to rest after traveling a total distance of $$7.90 m$$ along the ramp. Once stopped, a crate must not rebound back up the ramp.

Question:

Calculate the force constant of the spring that will be needed in order to meet the design criteria.

My answer is 27.4 N/m, but I am not quite sure that it is correct.

I know $$K_1=(1/2)(1500/9.80)(2.00^2)=306 J$$

$$K_2$$ is zero. $$U_1$$ is also zero. $$W_f$$ is $$-4270 J$$. I define $$x_2=7.90m$$.

So, I should use the equation $$K_1+U_1+W_f=mgy_2+(1/2)(k)(x_2)^2$$.

$$y_2=x_2*sin(24.0^\circ)=3.21$$. But, it is -3.21 because $$y_2$$ is below $$y_1=0$$.

Using the equation I believe I should use, I got the answer to be $$k=27.4 m/s.$$