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 crates weighing 1460 N will move at a speed of 1.80 m/s at the top of a ramp that slopes downward at an angle 21.0 degrees. The ramp exerts a kinetic friction force of 530 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 8.00malong 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.
Take free fall acceleration to be g= 9.80m/s^2.

Explanation / Answer

The energy the spring must absorb is ( taking g = 10 N/kg ) KE + GPE – Friction loses 1/2 . 146 . 1.8^2 + 1460 . 8 . sin 21 – 530 . 8 = 182.26 J For no rebound F(max) from spring must be 530 N so 530 = k . x 1/2 k x^2 = 182.26 530 . x = 364.52 x = 0.688 m k = 770.35 N/m

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