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?

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

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

Making x the subject of the formula, we have

Plugging in the numbers and solving for x, we get

Therefore, x = 0.05 m

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

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