A block with mass m = 6.4kg is hung from a vertical spring. When the mass hangs

Question

A block with mass m = 6.4kg is hung from a vertical spring. When the mass hangs in equilibrium, the spring stretches x= 0.27m. While at this equilibrium position, the mass is then given an initial push downward at v= 4.8 m/s; ignore friction.
a) what is the spring constant b) what is the oscillation frequency? c) After t= .032s, what is the speed of the block? d) what is the magnitude of the max acceleration of the block? e) at t=.32 s, what is the magnitude of the net force on the block? f) where is the potential energy of the system the greatest?

Please explain. Thank you. A block with mass m = 6.4kg is hung from a vertical spring. When the mass hangs in equilibrium, the spring stretches x= 0.27m. While at this equilibrium position, the mass is then given an initial push downward at v= 4.8 m/s; ignore friction.
a) what is the spring constant b) what is the oscillation frequency? c) After t= .032s, what is the speed of the block? d) what is the magnitude of the max acceleration of the block? e) at t=.32 s, what is the magnitude of the net force on the block? f) where is the potential energy of the system the greatest?

Please explain. Thank you.
a) what is the spring constant b) what is the oscillation frequency? c) After t= .032s, what is the speed of the block? d) what is the magnitude of the max acceleration of the block? e) at t=.32 s, what is the magnitude of the net force on the block? f) where is the potential energy of the system the greatest?

Please explain. Thank you.

Explanation / Answer

The spring constant K is computed with the information known about the mass at rest:

F = kx = m*g = k*0.27

k = m*g/0.27 = 6.4*9.81/0.27 = 232.534 N/m

The frequency of oscillation is:

f = sqrt( k/m ) / ( 2*pi ) = sqrt( 232.534 / 6.4 ) / ( 2*(22/7) ) = 0.9589 Hz

The kinetic energy at t = 0 is:

E = (1/2)*m*v^2 = (1/2)*6.4*(4.8)^2 = 73.728 J

At the extreme of motion, this translates entirely into additional spring potential energy. This point also represents the maximum acceleration.

Ep = (1/2)*k*(x)^2 = E

x = sqrt( 2*E / k ) = sqrt( 2*73.728 / 232.534 ) = 0.79632 m

The additional force of the spring is:

F = k*x = 232.534*0.79632 = 185.1715 N

F = m*a

a = F/m = 185.1714/6.4 = 28.933 m/s^2

a is the acceleration at maximum displacement, which is the maximum acceleration of the block, and so this is the answer of D part

The equation of motion of the block is then:

x = 0.27 + 0.79632*Sin( 2*pi*0.9589*t)

here we choose the Sin term for the motion, since the additional displacement is zero at t = 0.

The speed of the block is:

v(t) = dx/dt =0.79632*[ Cos( 2*pi*0.9589*t ) ]*(2*pi*0.9589)

v(.032) = 0.79632*6.02737*Cos( 6.02737* 0.032 ) = 4.7107 m/s { answer of c}

According to the equation of motion, the x displacement at 0.32 s is:

x(.32) = 0.27 + 0.79632*Sin( 6.02737*.32 ) = 1.0158435 m

This causes a spring force of:

F = k*x =232.534* ( 1.0158435 ) = 236.2182 N

PE = 0.5*kx2

if x is maximum then PE is maximum

maximum displacement (x_max) = 0.27 + 0.79632 = 1.06632 m

PE = 0.5*232.534*1.06632*1.06632 = 132.2 J

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