Question 1 A spring of spring constant k = 8.250 N/m is displaced form equilibri

Question

Question 1 A spring of spring constant k = 8.250 N/m is displaced form equilibrium by a distance of 0.1500 m. If the spring is strected form the displacement of 0.1500 m to a displacement of 0.3500 m, what is the change in spring potential energy between those two positions?
Question 2 A spring of spring constant k=12.50 N/m is hung vertically. A 0.500kg mass is then suspended from the spring. What is the displacement of the end of the spring due to the weight of the 0.500kg mass?
Question 3 A mass of 0.4000 kg is rasised by a vertical distance of 0.4500 in the earth's gravitational field. What in the change in its gravitational potential energy? This same mass is lowered by a vertical distance of 0.3500m in the earth's gravitational field. What is the change in its gravitational potential energy?

Explanation / Answer

change in energy is = 0.5*8.250* (.35^2 - .15^2) = 0.4125 J displacement is k*x = m*g ... x = m*g/k = 0.5*9.8/(12.5) = 0.392 m m*g*h = .4 * 9.8 *.45 = 1.76 J m*g*(change in h) = .4* 9.8 * (-.35) = -1.372J

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