Question

Suppose the spring constant or force constant of the spring is 32.0 N/m and a total of 2.0 kg hangs at the bottom. The mass is smacked upwards from its rest position so that it rises up by 0.40 m before it starts back down. What is the acceleration of the mass at either end of its motion

Answer 1

Step-by-step explanation:

Elongation in the spring when a mass of 2 kg hangs from it

k d = mg where d is the elongation

32 d = 2 x 9.8

d = .6125 m .

When it rises by .40 m , the elongation reduces to .6125 - .40 = .2125 m

upward restoring force at this point = 32 x .2125 = 6.8 N .

Its weight = 2 x 9.8 = 19.6 N acting downwards

Net force on mass of 2 kg = 19.6 - 6.8 N = 12.8 N .

acceleration in downward direction = net force / mass = 12.8 / 2 = 6.4 m/s²

At the bottommost point , total extension = .6125 + .40 = 1.0125 m

restoring force in upward direction = 32 x 1.0125 = 32.4 N .

Net force in upward direction = 32.4 - 19.6 N = 12.8 N

acceleration = 12.8 / 2 = 6.4 m /s² in upward direction .