Question

You are asked to design a spring that will give a 1020 kg satellite a speed of 2.25 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?

Answer 1

Answer:

(a) 2.45×10⁵ N/m

(b) 0.204 m

Explanation:

Here we have that to have a velocity of 2.25 m/s then the relationship between the elastic potential energy of the spring and the kinetic energy of the rocket must be

Elastic potential energy of the spring =  Kinetic energy of the rocket

$\frac{1}{2} kx^2 = \frac{1}{2} mv^2$

Where:

k = Force constant of the spring

x = Extension of the spring

m = Mass of the rocket

v =  Velocity of the rocket

Therefore,

$\frac{1}{2} kx^2 = \frac{1}{2} \times 1020 \times 2.25^2$

or

$kx^2 = 1020 \times 2.25^2 = 10,226.25\\So \ that \ the \ force \ on \ the \ satellite\ kx = \frac{10226.25}{x}$

(b) Since the maximum acceleration is given as 5.00×g we have

Maximum acceleration = 5.00 × 9.81 = 49.05 m/s²

Hence the force on the rocket is then;

Force = m×a = 1020 × 49.05 = ‭50,031 N

$kx = \frac{10226.25}{x} = 50031 \ N$

Therefore,

$x = \frac{10226.25}{ 50031} = 0.204 \ m$

(a) From which

$k = \frac{10226.25}{x^2} = \frac{50031}{x} = \frac{50031}{0.204} = 244,772.13 \ N/m$ or

Force constant of the spring, k = 2.45×10⁵ N/m.