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 2.0 m/s at the top of a ramp that slopes downward at an angle 25.0^circ. The ramp will exert a 625 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

total weight of the crates =1480 N mass of the crates m = 1480/9.8 =151.02 kg initial speed of the crates vA = 2 m/s frictional force f = 625 N angle =250 distance traveled by crates d = 7.8 m (include x) from conservation of energy(crates when just touches the spring), from A to B: 1/2mvA2+mgh1 =1/2mvB2+mgh2   (1/2)mvB2 = (1/2) mvA2 + mg(h1-h2)   ......... (1) from figure, h1 = (7.8) sin25 h2 = (x) sin25 h2 = (x) sin25 h1-h2 = (7.8-x)sin25 .......... (2) substitute the eq (2) in eq (1), we get (1/2)mvB2 = (1/2) mvA2 + mg (7.8-x) sin25 ........ (3) finally, from B to C :from conservation of energy 1/2mvB2+mgh2 = 0+1/2kx2     ....... (4) substitute the eq (3) in eq (4), we get substitute the eq (3) in eq (4), we get (1/2) mvA2 + mg (7.8-x) sin25 +mgxsin25 =1/2kx2 (1/2) mvA2 + mg (7.8) sin25 =1/2kx2 k = 10361.5/x2 (1/2) mvA2 + mg (7.8) sin25 =1/2kx2 k = 10361.5/x2 k = 10361.5/x2

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