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

Question

ou are designing a delivery ramp for crates containing exercise equipment. The crates of weight 1490 N will move with speed 1.9 m/s at the top of a ramp that slopes downward at an angle 25.0 . The ramp will exert a 629 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 8.0 m along the ramp; this distance includes x. Once stopped, a crate must not rebound back up the ramp.

Part A

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

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ou are designing a delivery ramp for crates containing exercise equipment. The crates of weight 1490 N will move with speed 1.9 m/s at the top of a ramp that slopes downward at an angle 25.0 . The ramp will exert a 629 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 8.0 m along the ramp; this distance includes x. Once stopped, a crate must not rebound back up the ramp.

Part A

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

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Explanation / Answer

When the crate comes to rest, the force of the spring must not be more than the effective weight of the crate plus static friction (otherwise the crate will rebound).

force on spring= force on slope+friction F=kx=mgsin(theta)+friction
F = kx = mgsin(theta) + friction = 1490sin(25) + 629 = 1258.7 N

Also by energy conservation

Elastic spring energy= kinetic energy+ gratitational potential energy - work done by friction
0.5*k*x^2=0.5*m*v^2+mgh-friction*distance

0.5kx^2=0.5mv^2+mgh+Fd
0.5kx^2 = 0.5*1490/9.81 * 1.92 + 1490*8*sin(25) - 629*8

0.5kx^2 = 274.15 + 5037.6 - 5032

kx^2 = 559.5

also kx = 1258.7 N

dividing both equation x = 0.444

which gives k = 2831.68 N/m

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