An object of mass 2kg is thrown vertically downwards with an initial kinetic energy of 100J. What is the distance fallen by the object at the instant when its kinetic energy has doubled?

To solve this problem, we need to use the principle of conservation of energy. The initial kinetic energy of the object is given as 100J. When the kinetic energy has doubled, it becomes 200J.

The increase in kinetic energy is due to the work done by gravity on the object as it falls. This work done is equal to the weight of the object times the distance it falls, which is mgh, where m is the mass of the object, g is the acceleration due to gravity, and h is the height (or distance fallen).

We can set up the equation as follows:

mgh = ΔK.E

where ΔK.E is the change in kinetic energy, which is 200J - 100J = 100J.

Substituting the given values:

2kg * 9.8m/s^2 * h = 100J

Solving for h gives:

h = 100J / (2kg * 9.8m/s^2) = 5.1m

So, the object has fallen a distance of approximately 5.1 meters when its kinetic energy has doubled.