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.

air exerts forces on falling objects near earth’s surface.

explanation:

the term 'free falls' refer to object that are falling towards the earth, in a situation in which gravity is the only force acting on the object.

for an object in free fall, the acceleration is constant (g, the gravitational acceleration), and so the velocity of the object constantly increases.

for objects near earth's surface, there is always another force acting on the object: the resistance due to the air. this force is opposed to the direction of fall of the object, and it is proportional to the velocity of the object, so as the object falls down, this force increases up to a point when it balances the force of gravity, and the object continues its motion at constant velocity (called terminal velocity). since in this case gravity is not the only force acting on the object, we are not in a situation of free fall.