Question

Consider a 0.00040 kg elastic with a uniform spring constant of 40.0 N/m. In order to launch the elastic directly upwards with a velocity of 16.0 m/s, how far from it’s equilibrium position would the elastic be stretched?

Answer 1

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

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

We see that the 1/2 cancels from both sides and we are left with

$kx^{2} = mV^{2}$

Making x the subject of the formula, we have

$x = \sqrt{\frac{mV^{2} }{k} }$

Plugging in the numbers and solving for x, we get

$x = \sqrt{\frac{(0.0004)(16)^{2} }{40} }$

Therefore, x = 0.05 m

The object has to be stretched by 0.05 m from its equilibrium position.