Question

A 0.55-kg object connected to a light spring with a force constant of 19.8 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.

m/s

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

m/s

(c) Determine the speed of the object as it passes the point 1.5 cm
from the equilibrium position.

m/s

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

m

Answer #1

a)applying law of conservation of energy,

maximum kinetic energy (has max. speed) = masimum potential energy stored in the spring(when it has max. compression)

or 0.5*0.55*v^2 = 0.5*19.8*0.04^2

or v=0.24 m/s

b)again conserving energy,

potential energy in the spring + kinetic energy of the objject= total energy

or 0.5*19.8*0.015^2 + 0.5*0.55*v^2 = 0.5*19.8*0.04^2

or v=0.222 m/s

c)when it passes the point 1.5 cm from the equilibrium point, it behaves as if it has an extension of 1.5cm

so the speed will be v=0.222 m/s

d)let the value of the compression be x.so,

conserving energy,

potential energy in the spring + kinetic energy of the objject= total energy

or 0.5*19.8*x^2 + 0.5*0.55*(0.5*0.24)^2 = 0.5*0.55*0.24^2

or x=0.0346 m

or x=3.46 cm

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