A 0.33 kg object connected to a light spring with a force constant of 21.6 N/m o

Question

A 0.33 kg object connected to a light spring with a force constant of 21.6 N/m oscillates on a frictionless horizontal surface. If the spring is compressed 4.0 cm and released from rest.

(a) Determine the maximum speed of the object.

0.323 m/s
(b) Determine the speed of the object when the spring is compressed 1.5 cm.
0.299 m/s
(c) Determine the speed of the object when the spring is stretched 1.5 cm.
0.299 m/s

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

I need help with this one....

Explanation / Answer

a). from energy conservation; Elastic P.E. of spring = K.E. of object => 0.5*k*x^2 = 0.5*m*v^2 => 21.6*(0.04)^2 = 0.33*v^2 => v = 0.323 m/s b). from energy conservation; Elastic P.E. of spring = K.E. of object => 0.5*k*[x1^2 - x2^2] = 0.5*m*v^2 => 21.6*[(0.04)^2 - (0.015)^2] = 0.33*v^2 => v = 0.299 m/s c). from energy conservation; Elastic P.E. of spring = K.E. of object => 0.5*k*[x1^2 - x2^2] = 0.5*m*v^2 => 21.6*[(0.015)^2 - (0.04)^2] = 0.33*v^2 => v = -0.299 m/s d). v = 0.323/2 = 0.1615 m/s from energy conservation; Elastic P.E. of spring = K.E. of object => 0.5*k*[x1^2 - x2^2] = 0.5*m*v^2 => 21.6*[(0.04)^2 - x^2] = 0.33*(0.1615)^2 => x = 0.0347 m

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