Help with work done by gravity

[ramin86]

A raindrop of mass 3.34 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 50 m, what is the work done on the raindrop by the graviatational force?
J
(b) What is the work done on the raindrop by air resistance?
J

For (a), I tried -mg delta y, to get -3.34x10^-5(9.8)(50), but the answer turned out to be wrong. Please help

[ehild]

A raindrop of mass 3.34 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 50 m, what is the work done on the raindrop by the graviatational force?

For (a), I tried -mg delta y, to get -3.34x10^-5(9.8)(50), but the answer turned out to be wrong. Please help

Why is it negative? Just recall, what is work?

ehild

[ramin86]

Well I found the formula somewhere, and plus it was going downward, but I guess it doesn't work out since the answer was wrong.

[ehild]

Well I found the formula somewhere, and plus it was going downward, but I guess it doesn't work out since the answer was wrong.

Work is force times magnitude of displacement times the cosine of the angle between them. Gravity points downward, the raindrop falls downward, so they make zero angle. The work done by the gravitational force is positive. If you give a negative number as result it will be wrong.

ehild

[ramin86]

Well I just multipled 3.34*10^-5(9.8)cos(0) to get 3.27e-4

Is this how I do the problem?

[ramin86]

Well I got A wrong, can anyone give me an explanation on how to do it?

And how do I do B?

[Pyrrhus]

Well let's start from scratch

Work Definition

W = \vec{F} \cdot \vec{r}

Dot Product:

\vec{A} \cdot \vec{B} = |\vec{A}||\vec{B}|cos
\theta

where \theta is the angle between them

\vec{A} \cdot \vec{B} = A_{x}B_{x} + A_{y}B_{y} +A_{z}B_{z}

Applying the definition to our problem we have

W_{gravity} = m \vec{g} \cdot \vec{r}

W_{gravity} = m |\vec{g}||\vec{r}|cos0^o

or

W_{gravity} = m(-g_{y})(-r_{y})

Answer for both cases is W = 0.016 J

Now Air Resistance Work



Applying again the definition to our problem we have

W_{air} = \vec{R} \cdot \vec{r}

W_{air} = |\vec{R}||\vec{r}|cos180^o

or

W_{air} = m(R_{y}})(-r_{y})

Do you know what R = ?