Spring help please

[math4me]

I am doing some revision for physics and cant evaluate these questions, can anybody help or give a formula.

Q1.

A mass of 2.2 Kg is suspended from a spring of spring constant K = 25 Nm^-1
What is the initial extension?

Q2.

The mass (in Q1) is pulled down a further 5cm and released.
What is the initial upward acceleration?

Q3.

Calaculate the velocity of the mass as it passes through the equilibrium position.

Q4.

What is the value of period?


Thanks so much :-)

[Chen]

Have you tried solving any of these questions yourself?

[math4me]

Im new to physics but yes i have tried

[Chen]

What have you tried using? :smile: At least the first question is very simple, but if you are not able to understand it or answer it, how do you expect to understand the other questions? I am not trying to put you down or anything. It's for your own good, really.

"Give a man a fish; you have fed him for today. Teach a man to fish; and you can sell him fishing equipment."

[ninaricci]

force= k*x
mg=kx
2*9.8=25*x
Initial extension is 0.784m
k*x-mg=m*acceleration
25*(0.784+0.05)-2*9.8=2*acceleration
acceleration=0.625m/s^2
work=force*distance cos angle
since the force and the displacement are in the same direction, then
work of spring=0.5*k*x^2-0.5kx^2
=0.5*25(0)-0.5*25*(.784+.05)=-10.425joule
work due to gravity is mgh-mgh=0-2*9.8*(.784+.05)=16.3joule
total work= 10.425+16.3=26.7joule
work=change in kinetic energy=.5(mV^2-mV^2)=.5(2V^2-0)
26.7=v^2
v=5.16m/s
period= 1/frequency
frequency=number of cycles/time
the number of cycles is not give in the example
i think it is supposed to be in the example
anyway i hope that this solution would be correct as i beleive.....actually think so
best wishes

Yes i have noticed that i used the mass value as 2 instead of 2.2 sorry