I have a Physics question about the work done on a raindrop as it falls vertically for a distance of (4.5)m.
The rain drop falls at a constant velocity of (1.8)ms^-1,
and has a mas of (7.2x10^-9)kg
I have to work out the Work Done.
I have the formula for Work Done as, W = F s cos (theta)
and using this I got the answer (3.65x10^-7).
But apparently the actual answer was (3.2x10^-7), and I was meant to use the formula
Work Done = mgh,
but I have never seen this formula before and I don't understand how the,
mass x the gravity x the height would = Work Done.
Can someone explain why Work Done = mgh, and how it can be applied to the question to give the answer (3.2x10^-7)?
The rain drop falls at a constant velocity of (1.8)ms^-1,
and has a mas of (7.2x10^-9)kg
I have to work out the Work Done.
I have the formula for Work Done as, W = F s cos (theta)
and using this I got the answer (3.65x10^-7).
But apparently the actual answer was (3.2x10^-7), and I was meant to use the formula
Work Done = mgh,
but I have never seen this formula before and I don't understand how the,
mass x the gravity x the height would = Work Done.
Can someone explain why Work Done = mgh, and how it can be applied to the question to give the answer (3.2x10^-7)?
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I understand the question (from a teaching-physics point of view) but the physics is ill-explained with this!
Anyway, the spirit of the exercise is to see that W = F s, s being the distance (i.e. the height h) and that the force on the drop is that of gravity (m g). Hence W = mg h. This answers your question on where m g h comes from. It is still force times displacement along the force line F=mg, s = h.
Truth is, there is no net force on the drop (which is why it moves at constant velocity). Hence the gravity and the air resistance force balance. Therefore there is no net work performed on the rain drop! (The earth's gravity field performs work on the drop, but the drop performs the same amount of work in pushing itself through the air...)
Ignoring this "finer point" and giving way to the "but i need a number..", the formula works out to
W = m g h
= 7.2 10^(-9) kg * 9.81 m/s^2 * 4.5 m
= 3.2 10^(-7) J
Again, this is the work done by the earths gravitational field on the drop.
On the point of your W=F s cos(theta). Your theta is 0 degrees (rain falls vertically so force and displacement are parallel). What did you use for F to get W = 3.65 10^(-7)...?
Anyway, the spirit of the exercise is to see that W = F s, s being the distance (i.e. the height h) and that the force on the drop is that of gravity (m g). Hence W = mg h. This answers your question on where m g h comes from. It is still force times displacement along the force line F=mg, s = h.
Truth is, there is no net force on the drop (which is why it moves at constant velocity). Hence the gravity and the air resistance force balance. Therefore there is no net work performed on the rain drop! (The earth's gravity field performs work on the drop, but the drop performs the same amount of work in pushing itself through the air...)
Ignoring this "finer point" and giving way to the "but i need a number..", the formula works out to
W = m g h
= 7.2 10^(-9) kg * 9.81 m/s^2 * 4.5 m
= 3.2 10^(-7) J
Again, this is the work done by the earths gravitational field on the drop.
On the point of your W=F s cos(theta). Your theta is 0 degrees (rain falls vertically so force and displacement are parallel). What did you use for F to get W = 3.65 10^(-7)...?