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 1810-N crates will move at 1.8 m/s at the top of a ramp that slopes downward at 22.0. The ramp exerts a515-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.

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

Explanation / Answer

here,

weight , w = 1810 N

mass , m = w/g = 184.69 kg

To avoid rebound, the maximum force of the compressed spring must not exceed the weight component of the crate acting parallel to the slope plus
the static friction force

F = kx = (mgsin(theta) + Ff)

x = (mgsin(theta) + Ff) / k

x = (1810*sin(22) + 515) / k

x = 1193.04/k

The maximum spring potential will equal the kinetic energy plus the change in potential energy minus the work done by friction from the top of the ramp

PS = KE + PE - W

0.5*k*x^2 = 0.5*mv^2 + mgh - Fd

0.5*kx^2 = 0.5*mv^2 + mgdsin(theta) - Fd

kx^2 = mv^2 + 2d(mgsin(theta) – F)

k(1193.04/k)^2 = mv^2 + 2d(mgsin(theta) – F)

1193.04^2 / k = mv^2 + 2d(mgsin(theta) – F)

k = 1193.04^2 / (mv^2 + 2d(mgsin(theta) – F))

k = 1193.04^2 / (184.69*(1.8^2) + 2(5.0)(1810*sin(22) – 515))

k = 638.62 N/m

the largest force constant of the spring that will be needed to meet the design criteria is 638.62 N/m

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