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UppityHorse
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Homework Statement
You are designing a delivery ramp for crates containing exercise equipment. The crates weighing 1490 N will move at a speed of 2.10 m/s at the top of a ramp that slopes downward at an angle 25.0°. 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 7.70 m along the ramp. Once stopped, a crate must not rebound back up the ramp.
Calculate the force constant of the spring that will be needed in order to meet the design criteria.
Homework Equations
I have been using:
[tex]K_{1} + U_{grav1} + U_{elastic1} - W_{friction} = K_{2} + U_{grav2} + U_{elastic2}[/tex]
The Attempt at a Solution
[tex]K_{1}= (1/{2})(1490/{9.8})(2.1^{2})=335.25 J[/tex]
[tex]U_{grav1} = U_{elastic1} = 0[/tex] (because at the top, I have y=0)
[tex]W_{friction} = (540)(7.7) = 4158 J[/tex] (force * distance)
[tex]K_{2} = 0[/tex]
[tex]U_{grav2} = (1490)(-7.7)(cos 25) = -10398.1 J[/tex]
[tex]U_{elastic2} = (1/2)(k)(7.7^{2})[/tex]
So, my first thought is, I have no idea how long this spring is, so I have no clue how much it is being compressed. Would I be correct to say that x=(7.7)^2 in the U_elastic equation?
Using this and solving for the k in the 2nd elastic equation, I am not getting a correct answer. I am getting 221.803.
This is what I input into my calculator, so you can see how I arranged the equation to come up with K:
(335.25-4158+10398.1)*2/(7.7^2)
Please point out my error, thanks a lot.