Question

A block of mass m = 1.5 kg is attached to a massless, frictionless vertical spring and stretches the spring by an amount y0 = 0.15m

a)find the spring constant k of the spring

b) the block is then pulled down by an additional 0.05m below its equilibrium position and is released. express the position of the block during its resulting simple harmonic motion using the equation y(t) = ymcos(wt+@).

c) find the maximum acceleration fo the block A(m).

d) find the total energy of the system of the block, spring and earth at the lowest point of the motion of the block. set the gravational potential energy of the block equal to zero at the midpoint of its motion

Answer 1

a) let spring constant be k

kx = mg

=> k *0.15 = 1.5 * 10

=> k = 100 N/m

b) ym = 0.05 m

angular velocity = k/m = 8.16 rad/s

equation = 0.05 cos(8.16 t)

c) acceleration = d²(0.05 cos(8.16 t)) / dt = (8.16)² * 0.05 cos(8.16 t)

max acceleration =   (8.16)² * 0.05 = 3.329 m/s²

d) spring potential energy = 1/2 k x² where x is total elongation = 0.15 + 0.1 = 0.25

= 1/2 * 100 * 0.25² = 3.125 J

gravitational potential energy = - m g h = - 1.5 * 10 * 0.05 =-0.75 J

kinetic energy of block at lowest poin t = 0 since velocity = 0

total energy =  spring potential energy + gravitational potential energy+ kinetic energy of block = 2.375 J