Two particles of the same mass m are moving in circular orbits because of force, given by $$F\left( r \right) = {{ - 16} \over r} - {r^3}$$

The first particle is at a distance r = 1, and the second, at r = 4. The best estimate for the ratio of kinetic energies of the first and the second particle is closest to :

A body of mass m starts moving from rest along x-axis so that its velocity varies as $$\upsilon = a\sqrt s $$ where a is a constant and s is the distance covered by the body. The total work done by all the forces acting on the body in the first t seconds after the start of the motion is :

An object is dropped from a height h from the ground. Every time it hits the ground it looses 50% of its kinetic energy. The total distance covered as t $$ \to $$ $$\infty $$ is :

A body of mass m = 10^–2 kg is moving in a medium and experiences a frictional force F = –kv². Its initial speed is v₀ = 10 ms^–1. If, after 10 s, its energy is $${1 \over 8}mv_0^2$$, the value of k will be: