Question

A 0.40 kg mass is attached to a spring with force constant 26 N/m and released from rest at a distance of 3.2 cm from the equilibrium position Use conservation of energy to find the speed of the spring when it is halfway to the equilibrium position Find the maximum speed of the mass How far is the mass from the equilibrium position when its speed is half the maximum speed?

Answer 1

3.2 cm from equilibrium position:
KE + PE = mg*d
0 + PE = mg*d/2

Halfway to the equilibrium position:
KE + PE = mg*d
mg*d/2 + mg*d/2 = mg*d

KE = mg*d/2 = 0.5m*V^2
0.5m*V^2 = mg*d/2 = 3.92*0.032/2
0.2*V^2 = 0.06272
V^2 = 0.3136
V = 0.56 m/s.

Answer 2

This is not correct. Use 1/2kx^2 for PE rather than mgd.

What you've done here doesn't incorporate the spring constant (k) which totally influences how fast the mass will be moving.

I got the answer .223 m/s when I used this equation.