Rupert finds that if he attaches a 430 gram mass to the hook, and then very slow

Question

Rupert finds that if he attaches a 430 gram mass to the hook, and then very slowly lowers the mass with his hand until the force on the mass by the spring is as large as the weight of the mass, that the spring stretches by 38 cm.
(a) What is the spring constant of Rupert's spring?
N/m

The 430 mass is now hanging from the end of the spring. Rupert grasps the mass with his hand and pulls it down by 18 cm. He then releases the mass from rest. You will be asked to find the speed of the mass after it has moved upwards by 18 cm from its release point. First, answer these preliminary questions.
What is the force on the mass by the spring at the instant after Rupert releases his grip on the mass?
N up
What is the weight of the hanging mass?
NDown
What is the net force on the mass at the instant after Rupert releases his grip on the mass?
NUp
And what is the acceleration of the mass at the instant after Rupert releases his grip on the mass?
m/s2Up

What is the work done on the mass by the spring as the mass moves the 18cm upward from its release point? What is the change in the elastic potential energy of the spring-masss-Earth system (the horizontal rod is fixed, eventually, to the Earth)as the mass the 18cm upward from its release point? What is the work done on the mass by gravity as the mass moves the 18cm upward from its release point? What is the change in the gravitational potential energy of the mass-Earth system as the mass moves the 18cm upward from its release point? What is the change in the total potential energy of the spring-mass-Earth system as the mass moves the 18cm upward from its release point? What is the speed of the mass after it has moved upwards by 18 cm from its release point?

Explanation / Answer

a) m = 0.43 kg
   x = 0.38 m

F = k*x
m*g = k*x

==> k = (m*g)/x = 11.0895 N/m

b)

F = k*x = 11.0985*(0.38+0.18) = 6.215 N

weight = m*g = 0.43*9.8 = 4.214 N

Fnet = F-weight = 2.001 N(upward)

a = Fnet/m = 4.653 m/s^2

finding velocity

initial mechanical enegy = final mechanical energy

0.5*k*x1^2 = m*g*x2 + 0.5*m*v^2 + 0.5*k*x2^2

here x1 = 0.38+0.18 = 0.56 m, x2 = 0.38

1.74 = 0.7585 + 0.5*0.43*v^2 + 0.8


v^2 = 0.84418

==> v = 0.9188 m/s

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