Question

A block with mass m =7.4 kg is hung from a vertical spring. When the mass hangs in equilibrium, the spring stretches x = 0.21 m. While at this equilibrium position, the mass is then given an initial push downward at v = 5.1 m/s. The block oscillates on the spring without friction.

3)After t = 0.4 s what is the speed of the block?

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

5)At t = 0.4 s what is the magnitude of the net force on the block?

Answer 1

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

F = kx = m*g = k*.2

k = m*g/.2 = 7.4*9.81/.21 = 345.6N/m

we agree on K!

The frequency of oscillation is:

f = sqrt( k/m ) / ( 2*π ) = sqrt( 345.6 / 7.4 ) / ( 2*π ) = 1.087 Hz

We agree here also!

The kinetic energy at t = 0 is:

E = (1/2)*m*v^2 = (1/2)*7.4*(5.1)^2 = 96.23 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*96.2 / 345.6 ) = .75 m

The additional force of the spring is:

F = k*Δx = 345.6*.75= 259.2 N

F = m*a

a = F/m = 259.2/7.4 =35.03 = 35m/s^2

a is the acceleration at maximum displacement, which is the maximum acceleration of the block, and so this is the answer to the second question.

The equation of motion of the block is then:

x = .21 + .75*Sin( 2*π*1.08*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 =.75*[ Cos( 2*π*1.08*t ) ]*(2*π*1.08)

v(0.4) = = -4.6 m/s

This means that the mass is moving upward at 4.6 m/s. Note that the argument of the Cos is in radians.

According to the equation of motion, the x displacement at 0.4 s is:

x(.4) = 0.21 + .75*Sin( 2π*1.08*0.4) = 0.520 m

This causes a spring force of:

F = k*x = 345.6 * ( .520) = 179.7 N

i.e. the spring is pulling the mass up. This figure includes the m*g weight of the mass.