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

Now we can use the velocity equation

v(t) = - A w sin (wt - phi) [eq 1}

w = sqrt(k/m) = 5.48 rad/s

Hmmmml...to find A, I guess we have to us conservation of energy,

(1/2) k x^2 = (1/2) m v^2

x^2 = v^2 * m/k

x = sqrt (v^2 * m / k)

x = A = 0.74814 m

Now plug everything into eq 1

v = - (0.7481) * (5.48rad/s) sin (5.48rad/s*0.3sec - pi/2)

v = 4.1m/s



4)

kA = ma,

a = kA/m

a = (294.3N/m) (0.7481/ 7.5kg)' . Kx=mg , k =mg/x

a = 29.4 m/s^2

5) the net force is the spring force at t = 0.3sec.

Fnet = k * x, so we just need x when t = 0.3sec.

x(t) = - A cos (wt - phi)

x = (0.7481m) cos (5.48rad/s*0.3sec - pi/2)

x = 0.62m

Fnet = k * x = 182.40 N

Potential energy will be highest at above equalibrium position.