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 ?

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CarliReifsteck CarliReifsteck · Answer 1 · 2020-10-06T06:30:32+02:00

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

[tex]U_{0}=Fx[/tex]....(I)

Constant force = 12F

A resistive force field is now set up ,

Resistive force is given by,

[tex]F_{r}=12F[/tex]

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

We need to calculate the kinetic energy

Using formula for kinetic energy

[tex]U=F_{r}x[/tex]

Put the value of [tex]F_{r}[/tex]

[tex]U=12Fx[/tex]

Now, from equation (I)

[tex]U=12U_{0}[/tex]

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