A block with mass m =7.0 kg is hung from a vertical spring. When the mass hangs

Question

A block with mass m =7.0 kg is hung from a vertical spring. When the mass hangs in equilibrium, the spring stretches x = 0.2 m. While at this equilibrium position, the mass is then given an initial push downward at v = 4.2 m/s. The block oscillates on the spring without friction. 1) What is the spring constant of the spring? 2) What is the oscillation frequency? 3) After t = 0.32 s what is the speed of the block? 4) What is the magnitude of the maximum acceleration of the block? 5) Where is the potential energy of the system the greatest? A) At the highest point of the oscillation. B) At the new equilibrium position of the oscillation. C) At the lowest point of the oscillation.

Explanation / Answer

The spring constant K is

F = kx = m*g = k * 0.2

k = m*g/0.2

= 7.0 * 9.81/0.2

= 343 N/m

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The frequency of oscillation is:

f = ( k/m ) / ( 2* )

= ( 343 / 7.0 ) / ( 2* )

= 1.11 Hz

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The kinetic energy at t = 0 is:

E = (1/2)*m*v^2

= (1/2) * 7.0 * (4.2)^2

= 61.74 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 )

= ( 2 * 61.74 / 343 )

= 0.6 m

The additional force of the spring is:

F = k * x

= 343 * 0.6

= 205.8 N

F = m*a

a = F/m

= 205.8 / 7

= 29.4 m/s^2

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