Question

You need to design a spring that will launch a 1060 kg satellite with a speed of 3.35 m/s relative to an orbiting space station. The maximum safe acceleration of the satellite is 5.00g, (g being the usual gravitational field constant on the surface of the Earth). The spring's mass, the recoil kinetic energy of the space station, and changes in gravitational potential energy will all be negligible.What must the force constant of the spring be

Answer 1

226.8 kN/m

Step-by-step explanation:

The work done by the spring, W equals the kinetic energy of the satellite, K

W = K

work done by the spring, W = 1/2kx² where k = force constant and x = extension of spring

kinetic energy of the satellite, K = 1/2mv² where m = mass of satellite = 1060 kg and v = speed of satellite = 3.35 m/s

1/2kx² = 1/2mv²

k = mv²/x²

Also, the spring force F = kx where k = force constant and x = extension of spring.

k = F/x

equation both expressions for k, we have

mv²/x² = F/x

x = mv²/F since F = ma where m = mass of satellite and a = maximum acceleration of satellite = 5.00g and g = 9.8 m/s²

x = mv²/ma = mv²/5.00mg = v²/5.00g

Substituting the values of the variables into the equation, we have

x = v²/5.00g

= (3.35 m/s)²/(5.00 × 9.8 m/s²)

= 11.2225 m²/s²/49 m/s²

=0.229 m

Now k = F/x = 5.00mg/x

Substituting the values of the variables into the equation, we have

k = 5.00mg/x

k = 5.00 × 1060 kg × 9.8 m/s²/0.229 m

k = 51940 kgm/s²/0.229 m

k = 51940 N/0.229 m

k = 226812.23 N/m

k = 226.81223 kN/m

k ≅ 226.8 kN/m