Particle Problem: Work Done on Raindrop by Grav & Air Resistance

[chocolatelover]

Homework Statement

A raindrop of mass 3.07 X 10^-5 kg falls vertically at constant speed under the influence of gravity and air resistance. Model the drop as a particle.

a. As it falls 50m, what is the work done on the raindrop by the gravitational force?

b. What is the work done on the raindrop by air resistance?

Homework Equations

The Attempt at a Solution

a. Fg=-mgj=(3.07X10^-5)(-9.8m/s^2j)
=-3.0086x10^-4

delta r=?

w=?

Could someone please tell me if this is correct so far and help me find delta r?

Thank you very much

 

[Shooting Star]

What do you mean by delta r?

Anyway, do you know the work done by gravity when a particle falls from a height h1 to a height h2?

Also, note that the drop is falling at a constant speed. So, the KE is not increasing, but the gravitational PE is decreasing.

Think for a while before rushing back here.

 

[chocolatelover]

What do you mean by delta r? Δr

Anyway, do you know the work done by gravity when a particle falls from a height h1 to a height h2? I know that w=FΔrcosθ

Also, note that the drop is falling at a constant speed. So, the KE is not increasing, but the gravitational PE is decreasing.

Could you please give me a hint? Is what I did so far correct?

Thank you

 

[nrqed]

Could you please give me a hint? Is what I did so far correct?

Thank you

\Delta \vec{r} is simply the final position minus the initial position so it's simply -50 m \hat{j} (where m here is for meters, it's not the mass). To get the work done by gravity, take the dot product of the force with the vector delta r

(another way of doing is is to take the difference of gravitational potential energy)

 

[Shooting Star]

Could you please give me a hint? Is what I did so far correct?

I specifically asked about the work done by gravity when a particle falls from a height h1 to a height h2, and you have blindly answered w=FΔrcosθ. To do this problem, you have to say what is the work done in terms of h1 and h2. Look at nrqed's post.

 

[chocolatelover]

Thank you very much

Could you tell me if this is correct?

a. As it falls 50m, what is the work done on the raindrop by the gravitational force?

Wouldn't this be 0J? Because when it is perpendicular the work is 0, right?

b. What is the work done on the raindrop by air resistance?

Fg=-mgj=(3.07X10^-5)(-9.8m/s^2)j
=-3.0086X10^-4=-.0003JN

Δr=initial-final

=0mi-50mj
=-50mj

w=(-50mj)(.003jN)
=.015W

Thank you

 

[Shooting Star]

Could you tell me if this is correct?

a. As it falls 50m, what is the work done on the raindrop by the gravitational force?

Wouldn't this be 0J? Because when it is perpendicular the work is 0, right?

It is not zero. I was not able to understand which is perpendicular to what. Please clarify.

b. What is the work done on the raindrop by air resistance?

Fg=-mgj=(3.07X10^-5)(-9.8m/s^2)j
=-3.0086X10^-4=-.0003JN

Δr=initial-final

=0mi-50mj
=-50mj

w=(-50mj)(.003jN)
=.015W

If you are taking the initial position to be (0,0), then,
Δr=initial-final = 0j - (-50)j. (What is the m doing there?)

w=(-50mj)(.003jN)
=.015W

This is, in fact, the work done by gravity on the body, although in this particular problem, the force due to resistance is equal to the force of gravity, because the drop is falling at a constant speed. The work done by gravity is dissipated into other forms of energy, mostly heat, due to the frictional force.

Work done by gravity = -(work done by air resistance).

 

[chocolatelover]

Thank you very much

Regards