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 1480 N will move with speed 2.2 m/s at the top of a ramp that slopes downward at an angle 22.0 degree The ramp will exert a 554 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 7.8 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 k_max that can be used in order to meet the design criteria.

Explanation / Answer

The component of the gravitational force acting along the ramp is 1480 * sin 22 which is 554 N

As the ramp exerts the frictional force of 554 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 ) 554 Gravity ( down) 554 spring (up) = 1108N
Which gives the net force of zero once the weight has stopped.

So the maximum force applied by the spring is 1108 N.

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 = Fx

x = m v^2 /F

m = 1480 / g
= 1480 /9.8
= 151 kg

x = 151 * 2.2^2 / 1108
= 0.66 m

And k = F/x
= 1108/ 0.66
= 1.6 * 10 ^ 3 N/m

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