A block of mass 3.00 kg is attached to an ideal spring placed on a horizontal, f

Question

A block of mass 3.00 kg is attached to an ideal spring placed on a horizontal, frictionless surface. The other end of the spring is fixed. A horizontal force of 25.0 N is required to hold the block at rest when it is pulled 0.400 m from its equilibrium position. The object is released from this position and undergoes SHM. What is the spring constant of the spring? What is the frequency (f) of the oscillations? What is the maximum kinetic energy of the block? Where does the block have its maximum kinetic energy? What is the maximum elastic potential energy of the system? Where does it occur? What are the kinetic and elastic potential energies of the system when its position is equal to one third of the maximum value? What are the speed and acceleration of the block when its position is equal to one fourth of the maximum value?

Explanation / Answer

vrms = sqrt(3RT/M) = 2017.4 m/s

a)

given F = 25 N

elastic force = F = k*x

k = F/x = 25/0.4 = 62.5 N/m

b)

f = (1/2pi)*sqrt(k/m)

f = (1/2pi)*sqrt(62.5/3)

f = 0.73 Hz

c)

KEmax = PEmax = 0.5*k*x^2 = 0.5*62.5*0.4^2 = 5 J

d)

at the minium position i.e at x = 0

e)

PE max = 0.5*k*x^2 = 0.5*62.5*0.4^2 = 5 J

f)

at the maximum displacement x = 0.4 m

g)

KE = 0.5*m*w^2*(A^2-x^2) = 0.5*k*(A^2-x^2)

x = A/3 = 0.4/3

KE = 0.5*62.5*(0.4^2-(0.4/3)^2)

KE = 4.44 J

PE = 0.5*k*x^2 = 0.5*62.5*(0.4/3)^2 = 0.56 J

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h)

v = w*sqrt(A^2-x^2) = 2*pi*f*(A^2-x^2)

acceleration a = -w^2*x

at x = A/4 = 0.4/4 = 0.1 m

v = 2*pi*0.73*sqrt(0.4^2-0.1^2)

v = 1.78 m/s

acceleration = a = w^2*A = (2*pi*0.73)^2*0.4

a = 8.415 m/s^2

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