You are designing a delivery ramp for crates containing exercise equipment. The

Question

You are designing a delivery ramp for crates containing exercise equipment. The 1490-N crates will move at 1.8 m/s at the top of a ramp that slopes downward at 22.0. The ramp exerts a 515-N kinetic friction force 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 5.0 m along the ramp. Once stopped, a crate must not rebound back up the ramp.

A. Calculate the largest force constant of the spring that will be needed to meet the design criteria.

Explanation / Answer

Work done by (gravity + friction + spring) = chaneg in kE

Wg+Wf+Ws = -0.5*m*u^2

1490*sin(22)*5 - 515*5 -Ws = -0.5*(1490/9.81)*1.8^2 = -246

Work done by spring is Ws = 0.5*k*x^2= 215.8+246 = 461.8 J

But spring force Fs = kx = 1490*sin(22) + 515 = 1073.16 N

x = 1073.16/k

then 0.5*k*(1073.16/k)^2 = 461.8

k = 1247 N/m

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