A 2.0 ð‘˜ð‘” package is released on a frictionless ramp inclined 53.1 ° above th

Question

A 2.0 ð‘˜ð‘” package is released on a frictionless ramp inclined 53.1 ° above the horizontal, 4.00 ð‘š from a long spring with force constant 120 ð‘/ð‘š that is attached to the bottom of the incline. The mass of the spring is negligible.

What is the speed of the package just as it reaches the spring?

What is the maximum compression of the spring?

Suppose instead that the ramp were not frictionless and the coefficient of kinetic

friction between the ramp and the package is ðœ‡ô°“ = 0.20. What is the maximum compression of the spring in this case?

Explanation / Answer

let x = max compression ... the distance down and parallel to the incline until max compression is (4 + x) let the point of max compression be zero for gravitational potential energy (GPE) ... so, the initial height of the block above 0 is h = (4 + x)sin(53.1) now ... conserve energy (initial GPE) - (work done by friction) = (final elastic potential energy) mgh - (f_k)(4 + x) = (1/2)kx^2 mg(4 + x)sin(53.1) - [u_k*mgcos(53.1)]*(4 + x) = (1/2)kx^2 the equation is quadratic in x ... since 0 is at the very bottom, x will be the positive root.

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