A 0.49-kg object connected to a light spring with a force constant of 21.2 N/m o

Question

A 0.49-kg object connected to a light spring with a force constant of 21.2 N/m oscillates on a frictionless horizontal surface. The spring is compressed 4.0 cm and released from rest.

(a) Determine the maximum speed of the object.
_______________cm/s

(b) Determine the speed of the object when the spring is compressed 1.5 cm.
______________cm/s

(c) Determine the speed of the object when the spring is stretched 1.5 cm.
_____________cm/s

(d) For what value of x does the speed equal one-half the maximum speed?
_____________ cm

Explanation / Answer

(a) Apply conservation of energy from maximum compression to maximum velocity position

kx2/2 = mv2/2; for x=0.04 cm

vmax= 0.26 ms-1

(b)Conservation of energy from starting point to x=1.5

kx12/2 + mv12/2= kx22/2+mv22/2 ; x1=-0.04, v1=0; x2=-0.015 m

v2=0.24 ms-1

(c) potential energy doesnt depend on positions sign

So, as before v2=0.24 ms-1

(d) now v=0.26/2=0.13

Apply conservation of energy as before from maximum kinetic position to v=0.13

x2=0.117 cm

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