(adsbygoogle = window.adsbygoogle || []).push({}); You are designing a delivery ramp for crates containing exercise equipment. The crates weighing [tex]1500 N[/tex] will move at a speed of [tex]2.00 m/s[/tex] at the top of a ramp that slopes downward at an angle [tex]24.0^\circ.[/tex] The ramp exerts a kinetic friction force of [tex]540 N[/tex]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 [tex]7.90 m[/tex] 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 [tex]K_1=(1/2)(1500/9.80)(2.00^2)=306 J[/tex]

[tex]K_2[/tex] is zero. [tex]U_1[/tex] is also zero. [tex]W_f[/tex] is [tex]-4270 J[/tex]. I define [tex]x_2=7.90m[/tex].

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

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

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

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# Homework Help: Calculating a force constant using kinetic and potential energy?

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