How Is Work Calculated for a Falling Raindrop?

[pringless]

A raindrop of mass 3.78 x 10-5 kg falls vertically at constant speed under the influence of gravity and air resistance. After the drop has fallen 114 m, what is the work done on the raindrop (a) by gravity? and (b) by air resistance?


do i start out with mgh?

 

[chroot]

Hi pringless,

First, recall that the definition of work is W = F d \cos{\theta}, where F is the applied force, d is the distance the object moved, and \theta is the angle between the force and distance the object moved. If you need to review this concept, please consult:

http://hyperphysics.phy-astr.gsu.edu/hbase/work2.html

The work done by gravity, as you suggested, is just W_g = F d \cos{\theta_g} = m g h. (Since the force due to gravity and the motion of the particle are in the same direction, and the cosine of 0 is 1).

Now, the particle is said to fall with a constant velocity. When you see the words "constant velocity," think to yourself "zero force." The raindrop has no total (net) force applied to it. It's simply moving along according to Newton's first law. Since the raindrop has no total force applied to it, it means that the force due to the air resistance must be equal in magnitude to the force due to gravity, but in the opposite direction. In other words, the air resistance pushes up, while gravity pulls down.

The work done by the air resistance is then W_a = F d \cos{\theta} = - m g h. The air resistance pushes up, while the drop falls down -- therefore, \theta_a is 180 degrees, and the cosine of 180 degrees is negative one.

Thus, the work done by air resistance is exactly the negative of the work done by gravity. The total work done on the raindrop is zero. Does this make sense? Well, as I've already said, the raindrop is moving with constant velocity, and thus has no net force applied to it at all. If there's no force, there's no work -- so this answer does make sense.

- Warren

 

[M98Ranger]

Yes, you can use the formula W = mgh to calculate the work done by gravity on the raindrop. Here, m represents the mass of the raindrop, g is the acceleration due to gravity (9.8 m/s^2), and h is the height through which the raindrop falls. So, the work done by gravity on the raindrop can be calculated as follows:

(a) W = (3.78 x 10^-5 kg)(9.8 m/s^2)(114 m) = 4.22 x 10^-2 J

This means that gravity has done 4.22 x 10^-2 joules of work on the raindrop as it falls through 114 meters.

To calculate the work done by air resistance, we need to consider the fact that the raindrop is falling at a constant speed. This means that the net force acting on the raindrop is zero, and the work done by air resistance must be equal and opposite to the work done by gravity.

(b) Since the work done by gravity is in the downward direction, the work done by air resistance must be in the upward direction. Therefore, the work done by air resistance can be calculated as:

W = -4.22 x 10^-2 J

This means that air resistance has done -4.22 x 10^-2 joules of work on the raindrop, which is equal in magnitude but opposite in direction to the work done by gravity. This negative sign indicates that the work done by air resistance is acting against the direction of motion of the raindrop.

In conclusion, the work done by gravity on the raindrop is 4.22 x 10^-2 joules, while the work done by air resistance is -4.22 x 10^-2 joules. Both these forces are equal in magnitude but opposite in direction, resulting in a constant speed of the raindrop as it falls through 114 meters.