You are designing a delivery ramp for crates containing exercise equipment. The
ID: 1417304 • Letter: Y
Question
You are designing a delivery ramp for crates containing exercise equipment. The 1530-N crates will move at 1.8 m/s at the top of a ramp that slopes downward at 22.0. The ramp exerts a 515-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.
Express your answer with the appropriate units.
Explanation / Answer
The component of the gravitational force acting along the ramp is 1530 * sin 22 which is 515 N
As the ramp exerts the frictional force of 515 N and the static friction has the same value, the speed of the crate will not increase going down the ramp. ( the net force is zero under these conditions)
After the spring is compressed the weight must not rebound so the net upwards force must be zero when the weight has stopped.
The friction will be attempting to prevent the weight being pushed back up the ramp.
In this case the forces are friction ( down ) 515 Gravity ( down) 515 spring (up) = 1030N
Which gives the net force of zero once the weight has stopped.
Hence the maximum force applied by the spring is 1030N.
The kinetic energy of the body is fully transferred to the spring when it has come to rest.
The change in gravitational potential of the body is exactly used in the friction so together they have no effect.
1/2 m v^2 = 1/2 F x ( average force * distance = work done.)
m v^2 = F x
x = m v^2 /F
m = 1530 / g = 1530 /9.8 = 156.22 kg
x = 156.22 * 1.8^2 / 1030 = 0.49 m
And k = F/x = 1030/ 0.49
= 2.10 * 10 ^ 3 N/m
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