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 1850-N crates will move at 1.8 m/s at the top of a ramp that slopes downward at 22.0 degree. 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

mass of the crate=weight/g

=1850/9.8=188.78 kg

let spring constant be k N/m and compression of the spring be x meters.

consider the point where the block finally stops to be the zero potential energy referrence point.

then initial height of the crate=5*sin(22)=1.873 m

using work - energy principle:

initial potential energy of the crate + initial kinetic energy of the crate -work done against friction

=potential energy of the spring

==>mass*g*initial height+0.5*mass*initial speed^2-friction force*distance =0.5*k*x^2

==>0.5*k*x^2=188.78*9.8*1.873+0.5*188.78*1.8^2-515*5=1195.956

==>k*x^2=2391.912 ....(1)

when the crate compresses the spring maximum,

spring force=static friction+component of weight along the ramp

==>k*x=515+1850*sin(22)=1208.022....(2)

dividing equation 1 by equation 2:

x=1.98 m

usign the value of x in equation 2,

k=610.1 N/m

so spring constant is 610.1 N/m

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