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You are asked to design a spring that will give a 1160-kg satellite a speed of 2.50 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?

Respuesta :

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Given:
m = 1160 kg
g = 9,81 m/s²
v = 2.5 m/s

Unknown:
k spring constant
x spring compression

1. equation to use is Newton's 2nd law and Hook's law:
[tex]F = ma = m(5g) = |kx|[/tex]
2. equation to use, the energy of the spring must equal the energy of the satellite:
[tex] \frac{1}{2} kx^2 = \frac{1}{2} mv^2 [/tex]
 Combinig the two equations:
[tex] \frac{1}{2} mv^2 = \frac{1}{2} m(5g)x \\ v^2 = 5gx \\ x = \frac{v^2}{5g} \\ \\ k = \frac{m(5g)}{x} = \frac{m(5g)^2}{v^2} [/tex]

Cricetus

The force constant and distance is:

  • (a) 29000 N/m
  • (b) 0.5 m

According to the question,

Mass,

  • m = 1160 kg

Speed,

  • v = 2.50 m/s

Acceleration,

  • a = 5.00 g

(b)

By applying the Hook law, we get

→ [tex]F = kx = ma[/tex]

By substituting the values, we get

               [tex]= 1160\times 2.50\times 5[/tex]

               [tex]= 14500[/tex]

Apply conversation of energy,

→  [tex]\frac{1}{2} mv^2 = \frac{1}{2} kx^2[/tex]

→     [tex]kx^2 = 1160\times (2.5)^2[/tex]

→     [tex]kx^2 = 7250[/tex]

→        [tex]x = \frac{7250}{14500}[/tex]

         [tex]x = 0.5 \ m[/tex]

(a)

The force constant will be:

[tex]= \frac{F}{x}[/tex]

[tex]= \frac{14500}{0.5}[/tex]

= [tex]29000 \ N/m[/tex]

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