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 1.7m/s at the top of a ramp that slopes downward at an angle 23.0^circ. The ramp will exert a 593N 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 7.8m 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

I'm going to modify the problem statement slightly, and then solve the modified problem. The modification is this: "Each crate will move a total distance of 7.7 m along the ramp; this distance does not include x". 1. Find the velocity of the crate at the bottom of the ramp. Net acceleration of crate = gravity - friction = g(sin 28.7 deg) - 721/1503 Net acceleration = a = (9.81)(0.480) - 0.480 a = 4.230 m/s/s v^2 = 2as = (4.23)(15.4) = 65.14 m^2/s^2 v = 8.071 m/s 2. Find the KE of the crate at the bottom of the ramp. This is the KE that the spring will convert to stored PE: KE = 0.5mv^2 KE = (751.5)(65.14) = 48,950 J 3. In addition to this stored PE, we need to place an additional constraint on the spring. We know that the parcel must not be forced back up the ramp. In other words: the maximum force of the spring, F = kx, must = [the parcel weight-component parallel to the ramp, plus the force of the static-friction force for the mass on the ramp]. Thus: F = kx = 721 N + mg(sin 28.7 deg) F = kx = 721 + 1503(9.81)(0.480) kx = 721 + 7077 = 7798 N 4. We now can write 2 equations with 2 unknowns: k, and x: U = 48950 J = 0.5kx^2 Let's solve for x: k = 7798/x; U = 48950 J = 0.5kx^2 U = 48,950 = 0.5(7798/x)(x^2) 48,950 = 3899x x = 48,950/3899 = 12.55 meters k = 7798/12.55 = 621 N/m ---------------- And now we see how my modification at the beginning -- "Each crate will move a total distance of 7.7 m along the ramp; this distance does NOT include x." -- created a solvable problem. The more difficult problem will, as they say, "be left as an exercise for the student".

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