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 of weight 1500 N will move with speed 1.8 m/s at the top of a ramp that slopes downward at an angle 21.0 . The ramp will exert a 537 N force of kinetic friction on each crate, and the maximum force of static friction also has this value. At the bottom of the ramp, each crate will come to rest after compressing a spring a distance x. Each crate will move a total distance of 7.6 m along the ramp; this distance includes x. Once stopped, a crate must not rebound back up the ramp.

Calculate the maximum force constant of the spring kmax that can be used in order to meet the design criteria.

Explanation / Answer

by conservation of energy

initial energy = final energy

mgh + 0.5 * mv^2 = 0.5 * kx^2 + friction force * distance

1500 * 7.6 * sin(21) + 0.5 * 1500 * 1.8^2 / 9.8 = 0.5 * k * x^2 + 537 * 7.6

at bottom the crate is in equilibrium so net force on it will be 0

kx = friction force

kx = 537

on solving we'll get

k = 571.812 N/m

maximum force constant of spring = 571.812 N/m

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