Force constant of a spring launching a satellite

  • Thread starter ph123
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  • #1
ph123
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You are asked to design a spring that will give a 1200 kg satellite a speed of 3.80 m/s relative to an orbiting space shuttle. Your spring is to give the satellite a maximum acceleration of 5.00g. The spring's mass, the recoil kinetic energy of the shuttle, and changes in gravitational potential energy will all be negligible.

a) What must the force constant of the spring be?

b) What distance must the spring be compressed?

I don't really know how to approach this problem, since without the spring constant (k), you cannot find the distance the spring is compressed (x) and vice versa.

But, according to the problem,

0.5kx^2 + mgh = 0.5mv^2 +5mgh
0.5kx^2 = 0.5mv^2 + 4mgh
kx^2 = mv^2 + 8mgh
k = (mv^2 + 8mgh)/x^2

I know this has to be wrong because there are three unknowns, h, x, and k. Can someone help me approach this differently?
 

Answers and Replies

  • #2
Dick
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You were told to ignore the gravitational potential energy, so drop the mgh terms. Then your energy equation is correct, but you still have two unknowns. Now use the condition that the maximum acceleration the spring is supposed to impart is 5g. (F=ma etc).
 
  • #3
ph123
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0.5mv^2 = 0.5kx^2

I am not seeing how to incorporate acceleration into this equation. To find the work done by the force of the spring,

F= ma = 5mg * x = 5mgx ??
 
  • #4
Dick
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For a spring, F=k*x, right? You don't incorporate it into that equation, you derive another equation. 2 equations+2 unknowns=happiness.
 
  • #5
ph123
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ahhhh.

so

F=kx
5mg=kx
x=(5mg)/k

0.5mv^2 = 0.5kx^2
mv^2 = kx^2
mv^2 = k([5mg]/k)^2
mv^2 = k(25m^2g^2)/(k^2)
k = (25mg^2)/v^2

right? i think so, yay
 

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