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 25.0 . The ramp will exert a 642 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.

Explanation / Answer

Refer above figure,

Use equation,

Work done by crate = KE + PE

W = KE + PE

W= Fk*d = - 642*7.8 = -5007.8 J

KE = 1/2m(vf^2-vi^2) = 1/2*(W/g)*(0^2- 2.2^2) = 1/2*(1520/9.8)(0-1.8^2) = -251.3 J

PE = 1/2kx2 –mgh = 1/2kx2 – 1520*8sin25 = 1/2kx2 – 5139.01 J

W = KE + PE

-5007.8 J = -251.3 J + 1/2kx2 – 5139.01 J

kx2 = 765.02 ---------------(1)

Now Fs = mgsinq+Fk = 1520sin25 + 642 = 1284.2 N

Also Fs= k*x= 1284.2 ----------------(2)

From (1) and (2)

k= (kx)^2/kx^2 = (1284.2)^2/765.02 = 2155.7 N/m

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