A block with mass m =7.3 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 = 4.5 m/s. The block oscillates on the spring without friction.
At t = 0.35 s what is the magnitude of the net force on the block?
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 = 6.8*9.81/.2 = 333.5 N/m
we agree on K!
The frequency of oscillation is:
f = sqrt( k/m ) / ( 2*π ) = sqrt( 333.5 / 6.8 ) / ( 2*π ) = 1.11 Hz
Our stating point is pi/2, going down from the equilibrium position.
Now we can use the velocity equation
v(t) = - A w sin (wt - phi) [eq 1}
w = sqrt(k/m) = 6.53 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 = 0.414 m
Now plug everything into eq 1
v = - (0.414 m) * (6.53 rad/s) sin (6.53 rad/s*0.39 sec - pi/2)
v = 2.24 m/s
EDIT -- I changed part 3 a bit.
4) You know, I was going to use the a(t) equation, but we don't know t where the acceleration is max (where x = A), so we would use the x(t) formula to find the time .......
But try this, when x = A the sum of the forces = the spring force = kA and also equals ma, so
kA = ma,
a = kA/m
a = (306.78N/m) (0.414 m) / 7.2 kg
a = 17.6 m/s^2
5) the net force is the spring force at t = 0.39 sec.
Fnet = k * x, so we just need x when t = 0.39 sec.
x(t) = - A cos (wt - phi)
x = (0.414 m) cos (6.53 rad/s*0.39 sec - pi/2)
x = 0.343 m
Fnet = k * x = 105.19 N