A 1.0-kg mass is suspended vertically from a spring with k = 113 N/m and oscilla
ID: 1693877 • Letter: A
Question
A 1.0-kg mass is suspended vertically from a spring with k = 113 N/m and oscillates with an amplitude of 0.4 m. At the top of its oscillation, the mass is hit in such a way that it instantaneously moves down with a speed of 1.4 m/s.(a) Determine its total mechanical energy.
________
(b) Determine how fast it is moving as it crosses the equilibrium point.
__________
(c) Determine its new amplitude.
________
the answers i have are ... but they are all wrong
a) 8.06 J
b) 4.015 m/s
c) 0.378 meters
thank you for all your help
A 1.0-kg mass is suspended vertically from a spring with k = 113 N/m and oscillates with an amplitude of 0.4 m. At the top of its oscillation, the mass is hit in such a way that it instantaneously moves down with a speed of 1.4 m/s.
(a) Determine its total mechanical energy.
________
(b) Determine how fast it is moving as it crosses the equilibrium point.
__________
(c) Determine its new amplitude.
________
the answers i have are ... but they are all wrong
a) 8.06 J
b) 4.015 m/s
c) 0.378 meters
thank you for all your help
(a) Determine its total mechanical energy.
________
(b) Determine how fast it is moving as it crosses the equilibrium point.
__________
(c) Determine its new amplitude.
________
the answers i have are ... but they are all wrong
a) 8.06 J
b) 4.015 m/s
c) 0.378 meters
thank you for all your help
Explanation / Answer
Amplitude A = 0.4 m Spring constant K = 113 N/m mass m = 1 Kg (a) Total mechanical energy E = (1/2)KA^2 + (1/2)m(v')^2 = 0.5 * 113 * (0.4)^2 + 0.5 * 1 * (1.4)^2 = 9.04 + 0.98 = 10.02 J (b) Velocity at mean position v = ? E = (1/2)mv^2 + (1/2)m(v')^2 10.02 = 0.5 * 1 * v^2 + 0.98 v^2 = 18.08 v = 4.252 m/s (c) Let A' be the new amplitude. (A')^2 = mv^2/K A' = Sqrt[mv^2/K] = Sqrt[0.16] = 0.4 m Amplitude A = 0.4 m Spring constant K = 113 N/m mass m = 1 Kg (a) Total mechanical energy E = (1/2)KA^2 + (1/2)m(v')^2 = 0.5 * 113 * (0.4)^2 + 0.5 * 1 * (1.4)^2 = 9.04 + 0.98 = 10.02 J (b) Velocity at mean position v = ? E = (1/2)mv^2 + (1/2)m(v')^2 10.02 = 0.5 * 1 * v^2 + 0.98 v^2 = 18.08 v = 4.252 m/s (c) Let A' be the new amplitude. (A')^2 = mv^2/K A' = Sqrt[mv^2/K] = Sqrt[0.16] = 0.4 m Amplitude A = 0.4 m Spring constant K = 113 N/m mass m = 1 Kg (a) Total mechanical energy E = (1/2)KA^2 + (1/2)m(v')^2 = 0.5 * 113 * (0.4)^2 + 0.5 * 1 * (1.4)^2 = 9.04 + 0.98 = 10.02 J (b) Velocity at mean position v = ? E = (1/2)mv^2 + (1/2)m(v')^2 10.02 = 0.5 * 1 * v^2 + 0.98 v^2 = 18.08 v = 4.252 m/s (c) Let A' be the new amplitude. (A')^2 = mv^2/K A' = Sqrt[mv^2/K] = Sqrt[0.16] = 0.4 mRelated Questions
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