Question

The magnitude of the force associated with the gravitational field is constant and has a value FF . A particle is launched from point BB with an initial velocity and reaches point AA having gained U0U0 joules of kinetic energy. A resistive force field is now set up such that it is directed opposite the gravitational field with a force of constant magnitude 12F12F . A particle is again launched from point BB . How much kinetic energy will the particle gain as it moves from point BB to point AA ?

Answer 1

Answer:

The kinetic energy of the particle will be 12U₀

Explanation:

Given that,

A particle is launched from point B with an initial velocity and reaches point A having gained U₀ joules of kinetic energy.

Constant force = 12F

According to question,

The kinetic energy is

$U_{0}=Fx$....(I)

Constant force = 12F

A resistive force field is now set up ,

Resistive force is given by,

$F_{r}=12F$

When the particle moves from point B to point A then,

We need to calculate the kinetic energy

Using formula for kinetic energy

$U=F_{r}x$

Put the value of $F_{r}$

$U=12Fx$

Now, from equation (I)

$U=12U_{0}$

Hence, The kinetic energy of the particle will be 12U₀.