A 1.0-kg mass is suspended vertically from a spring with k = 113 N/m and oscilla

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 m

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