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 1520 N will move with speed 1.8 m/s at the top of a ramp that slopes downward at an angle 24.0 degrees. The ramp will exert a 618 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.
Kmax=?N/m

Explanation / Answer

the mass of the crate m = F / a
the height of fall of the crate = 7.6 m *sin(24.0)
the velocity by the time it reaches the bottom of the ramp
v = sqrt2gh
the kinetic energy that is gained by the crate during the fall
K = mv^2/ 2
m = mass of the crate
v = velocity of the crate
the energy lost by the crate due to the friction
E = F*S
F = fricitonal force
S = distance moved on the inclined plane
hence the energy of the crate by the time it compresses the spring
K - E = W
W = is the work done in compressing
If the crate stops after compressing the spring to a distance of x then the potential
energy
kx^2 / 2 = W

k = force constant
W = kintic energy of the crate applied incompressing the spring.

Hope this helps!

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