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do i start out with mgh?

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- Thread starter pringless
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- #1

- 43

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do i start out with mgh?

- #2

chroot

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Hi pringless,

First, recall that the definition of work is [itex]W = F d \cos{\theta}[/itex], where [itex]F[/itex] is the applied force, [itex]d[/itex] is the distance the object moved, and [itex]\theta[/itex] 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 [itex]W_g = F d \cos{\theta_g} = m g h[/itex]. (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 [itex]W_a = F d \cos{\theta} = - m g h[/itex]. The air resistance pushes up, while the drop falls down -- therefore, [itex]\theta_a[/itex] 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

First, recall that the definition of work is [itex]W = F d \cos{\theta}[/itex], where [itex]F[/itex] is the applied force, [itex]d[/itex] is the distance the object moved, and [itex]\theta[/itex] 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 [itex]W_g = F d \cos{\theta_g} = m g h[/itex]. (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 [itex]W_a = F d \cos{\theta} = - m g h[/itex]. The air resistance pushes up, while the drop falls down -- therefore, [itex]\theta_a[/itex] 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

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