A mass (M = 3kg) is dropped off of a cliff. It lands on a spring with spring con

Question

A mass (M = 3kg) is dropped off of a cliff. It lands on a spring with spring constant k = 2.500 N/m a distance h = 30 below. What is the speed of the mass when it hit the spring? v = Give the gravitation potential and kinetic energy of the mass, and the potential energy of the spring, at the point where the spring is its most compressed (a distance Delta y), in terms of g. Delta y. k. and m. Ui = Us = KE = What is the distance Delta y? Delta y = How much work is done by the spring from the moment the mass hits the spring to when the spring is fully compressed? Wspring = If die mass bounces off of the spring up to maximum height which is 7 meters below is original drop point, how much mechanical energy is lost? Energy Lost =

Explanation / Answer

A) ok so we're going to assume that h = 0 at the top of the spring and its being dropped from rest (0 KE). Therefore: (1/2)mv^2 = mgh (1/2)v^2 = gh v = sqrt(2gh) v = sqrt(2*9.8*30) v = 588 m/s B) Us = (1/2)ky^2 Ug = mgy (these are just definitions) When the spring is most compressed, that means the object has breifly stopped moving and therefore its KE = 0. C) Now, because this is an energy conserving system, the initial energy (on the left side of the following equation) is equal to the final energy (y is the distance below h=0; the stuff on the right is the Us + Ug) mgh = (1/2)ky^2 + mgy (3kg)(9.8)(30m) = (1/2)(2500 N)y^2 + (3kg)(9.8)y 882 = 1250y^2 + 29.4y (ok the algebra is a little hard here so I just plugged in some random values for y and then adjusted it until I found the answer but if you are required to show your work then just give it a shot.) y = .8283 m D) W = Force*displacement W = (1/2)(2500)(.8283^2)*(.8283) W = 710.35 J E) Mechanical energy at original drop point was 882 J. The energy at 7 meter below is (3kg)(9.8)(23 m) = 676.2 J. The difference between them is 205.8 J (this is the energy lost).

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