A block with mass m =7.6 kg is hung from a vertical spring. When the mass hangs

Question

A block with mass m =7.6 kg is hung from a vertical spring. When the mass hangs in equilibrium, the spring stretches x = 0.29 m. While at this equilibrium position, the mass is then given an initial push downward at v = 4.4 m/s. The block oscillates on the spring without friction.

spring constant=257.0897N/m

oscillation frequency=.913599Hz

1. After t = 0.36 s what is the speed of the block?

-on this questioned I tried

V= -4.4sin(5.74*0.36+pi/2)

=-4.4sin(3.636)

=-4.4*-0.474

= 2.088 m/s

and this answer was incorrect and I don't understand how

2. At t = 0.36 s what is the magnitude of the net force on the block?

Explanation / Answer

For part (1), your process is correct and answer is also correct. The same process I have also tried.

For Part (2), 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 = 0.756 m

the net force is the spring force at t = 0.36 sec.

Fnet = k * x, so we just need x when t = 0.36 sec.

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

x = (0.756 m) cos (5.74 rad/s*0.36 sec - pi/2)

x = 0.665 m

Fnet = k * x = 170.90 N

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