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Physics problem about springs PLEASE HELP!

  • Thread starter dardar22
  • Start date
A 2 kg block hangs from a spring. A 253g body hung below the block stretches it a further 2.5cm. What's the spring constant? Use g=9.8N/kg. Calculate the value to 1 decimal place and use N/m as your unit.

A 1.8kg block hangs from a spring. A 221g body hung below the block stretches it a further 3.8cm. If the 221g body is removed and the block set into oscillation, what is the period of the motion? Use g=9.8N/kg. Calculate the value to 2 decimal place and use second as your unit.

Answers and Replies

In order to get help, you should post an attempted solution.

Here is a hint: Write the force balance equation for a) just the first block and b) both blocks together. How can you then use these two equations to obtain an expression for k in terms of the known values?

for the second part, if I were to do it, I would set up a force balance equation (left side in terms of displacement x from the rest postion, right side in terms of acceleration = d^2x/dt^2) with initial conditons dx/dt(0)=0 and x(0)=-x1. This could then be solved, and the period of the solution can be obtained.
What a know/think!

number 1??
I THINK.. it is
we know F=KX & W=MG
since F=W, KX=MG
so K*d=2kg*G also K*(d+2.5cm)=2.253kg*G
so K=(2kg*G)/d also K=(2.253kg*G)/(d+2.5cm)
then (2kg*G)/d = (2.253kg*G)/(d+2.5cm)
we solve for d: (d+2.5cm)/d= (2.253kg*G)/(2kg*G)
G cancels out: (d+2.5cm)/d= 2.253kg/2kg
so 1+(2.5cm/d)= 2.253/2
2.5cm/d= 2.253/2-1
d= 2.5cm/(2.253/2-1)=
now that you have d and d+2.5cm
you can plug into equation F=KX or K=F/X which is K=MG/X
use K=MG/X
K=2kgG/d you know G and d so you have K which is spring constant.

Number 2..