A 5.0-kg object is attached to a spring and placed on a horizontal, frictionless

Question

A 5.0-kg object is attached to a spring and placed on a horizontal, frictionless surface. A horizontal force of 50.0 N is required to hold the object in place at rest when it is pulled 0.20 m from its equilibrium position. The object is then released from rest from its stretched position, and undergoes simple harmonic motion. a. What is the frequency of the oscillations? b. What is the maximum speed of the object? c. What is the energy the oscillating system? d. What are the position and velocity of the object when half the energy is kinetic energy? e. What is the equation for the position of the object as a function of time?

Explanation / Answer

We know that F =-kx

k = 50 / 0.20 = 250 N/m

(a) we know that w = sqrt(k/m)

w = sqrt(250/5) =7.071 rad/s

w= 2pi*f

hence f = w/2pi =7.071/2/3.14 = 1.125 per sec.....Ans.

(b) F = ma

a = 50/5 = 10 m/s2

v2 = u2 + 2*a*s

v2 = 0 + 2*10*0.2 = 4

v = 2.0 m/s....Ans.

(c) total energy = 0.5kx2 =0.5*250*0.22 = 5 joule.....Ans.

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