# physics

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?

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1. Do your own work and stop looking on the Internet for answers.

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2. GET A LIFE CHUCK

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3. Chuck, there's nothing wrong with asking for help on the Internet especially if someone lost. They, shouldn't expect someone to solve the problem for them. nor should the "tutor" give them the answer but explain the process and thinking why we use what eq. but yes just giving them the answer won't help them learn "how" "when" or "Why". but asking is better than staring at a wall of text and numbers they don't understand or trying guess and learning it wrong.

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4. In order to solve this, you first have to find the displacement at time 0.35, so using the equation x(t) = Asin(wt-shift), and then by using F=-kx you are able to solve for the net force on the block at that time

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5. 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

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