Question

A block with mass m =7.5 kg is hung from a vertical spring. When the mass hangs in equilibrium, the spring stretches x = 0.25 m. While at this equilibrium position, the mass is then given an initial push downward at v = 4.1 m/s. The block oscillates on the spring without friction.

After t = 0.3 s what is the speed of the block?

What is the magnitude of the maximum acceleration of the block?

At t = 0.3 s what is the magnitude of the net force on the block?

Where is the potential energy of the system the greatest?

At the highest point of the oscillation.

At the new equilibrium position of the oscillation.

At the lowest point of the oscillation.

Answer 1

First of all determine the spring constant (k) of the spring.

Use the expression -

F = k*x = m*g = k*0.25

=> k = m*g/0.25 = (7.5*9.81)/0.25 = 294.3 N/m

So, the frequency of oscillation is:

f = sqrt( k/m ) / ( 2*π ) = sqrt( 294.3 / 7.5) / ( 2*π ) = 1.0 Hz

Now the mass is given an initial push of v = 4.1 m/s in the downward direction.

So, the kinetic energy at t = 0 is:

E = (1/2)*m*v^2 = (1/2)*7.5*(4.1)^2 = 63.04 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*(Δx)^2 = E

Δx = sqrt( 2*E / k ) = sqrt( 2*63.04 / 294.3) = 0.65 m

So, the additional force of the spring is:

F = k*Δx = 294.3 * 0.65 = 191.3 N [ Answer of Part (c)]

Now -

F = m*a

a = F/m = 191.3 / 7.5 = 25.5 m/s^2 [Answer of Part (b)]

Where, ‘a’ is the acceleration at maximum displacement.

Again, the equation of motion of the block is then:

x =0.25 + 0.65*Sin( 2*π*1.0*t)

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 = 0.65*[ Cos( 2*π*1.0*t ) ]*(2*π*1.0) = 4.08*[ Cos( 2*π*1.0*t ) ]

at t = 0.3 s

v(0.3) = 4.08 *Cos(2*π*1.0*0.3) = -1.26 m/s [Answer of Part (a)]

Part (d) –

The potential energy of the system will be greatest at the highest point of the oscillation.