We know from the question that the Mass of the elastic object m = 0.00040 kg
Its Spring Constant is given as k = 40.0 N/m
Velocity required when the object is launched V = 16.0 m/s
Distance the elastic object has to be stretched from the equilibrium x = ?
We can use Conservation of Energy in this case, which tells us that
Elastic Potential Energy stored in the elastic object = Kinetic energy gained by it when released
[tex]\frac{1}{2} kx^{2} = \frac{1}{2} mV^{2}[/tex]
We see that the 1/2 cancels from both sides and we are left with
[tex]kx^{2} = mV^{2}[/tex]
Making x the subject of the formula, we have
[tex]x = \sqrt{\frac{mV^{2} }{k} }[/tex]
Plugging in the numbers and solving for x, we get
[tex]x = \sqrt{\frac{(0.0004)(16)^{2} }{40} }[/tex]
Therefore, x = 0.05 m
The object has to be stretched by 0.05 m from its equilibrium position.
