A 0.50kg mass is attached to a spring with a force constant of 22N/m and release

Question

A 0.50kg mass is attached to a spring with a force constant of 22N/m and released from rest a distance of 4.0cm from the equilibrium position of the spring. A)What is the maximum speed of the mass? B)How far is the mass from the equilibrium position when its speed is half the maximum speed? Two sig figs for both answers A 0.50kg mass is attached to a spring with a force constant of 22N/m and released from rest a distance of 4.0cm from the equilibrium position of the spring. A)What is the maximum speed of the mass? B)How far is the mass from the equilibrium position when its speed is half the maximum speed? Two sig figs for both answers

Explanation / Answer

V max = A*sqrt(k/m) = .04*sqrt(2/.5)= .26533 m/s

.5*k*A^2 = .5*k*x^2 + .5*m*(v max /2)^2

=> x = 3.46 cm past the equillibrium point

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