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 1520N will move with speed 2.0m/s at the top of a ramp that slopes downward at an angle 22.0?. The ramp will exert a 569N 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 7.6m 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. in N/m

Explanation / Answer

Mass of crate, m= W/g = 1520/9.8 = 155.1 kg

Energy lost = Energy gained

1/2 mv^2 + mgh = f*s + 1/2 kx^2

310.2+4327.46 = 569*7.6 + 1/2 kx^2

kx^2 = 313.24

When the crate compresses the spring, force acted by the spring up the ramp = kx

Weight component down the ramp = 1520sin22 = 569.4 N

Max static friction down the ramp = 569N

So, max kx can be = 569+569.4 = 1138.4 N

kx<1138.4

From the two expressions, x>0.275 m

k<4137.3 N/m

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