# Homework Help: Help with work done by gravity

1. Oct 24, 2004

### 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

2. Oct 24, 2004

### ehild

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

ehild

3. Oct 24, 2004

### 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.

4. Oct 24, 2004

### ehild

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

5. Oct 24, 2004

### 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?

6. Oct 24, 2004

### ramin86

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

And how do I do B?

7. Oct 24, 2004

### 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 $R = ?$

Last edited: Oct 24, 2004