After reaching terminal velocity, a raindrop of m=3.06E-4kg falls vertically down a distance of 500m.
a)what is the work done by the force of gravity on the raindrop?
sol) since the displacement is in the same direction as the force, the work done is Fcos90 = 0. (correct?)
b) what is the work done by the force of air resistance on the raindrop?
sol) again, 0. I have big doubts on both answers because it moves in same direction. please help.
Another interesting question: in terms of unit vectors i or j. I know that the normal force of a particular object is mgcostheta (i hat) - fsintheta (j hat). My question is, since the normal force is always vertical, should I only report -fsintheta (j hat) as the only normal force acting on a skier in terms of unit vectors?
work done by gravity on the raindrop is mgh.
work done by air resistance=Force*distance, but the force is equal to mg, in the upward direction, so work done by friction is
-mgh
Notice the net work (gravity +friction) is zero, that is why the KE does not increase.
a) The work done by the force of gravity on the raindrop can be calculated using the formula: work = force * distance * cosine(theta). In this case, the force of gravity is acting vertically downward, which means the angle between the force and the displacement is 0 degrees. So the cosine of 0 degrees is 1, and the work done by gravity on the raindrop is given by: work = force * distance * cosine(0) = force * distance = m * g * d, where m is the mass of the raindrop, g is the acceleration due to gravity, and d is the distance fallen.
b) The work done by the force of air resistance is dependent on the nature of air resistance. In the case of a raindrop falling at terminal velocity, the force of air resistance is equal in magnitude but opposite in direction to the force of gravity. This means that the net force acting on the raindrop is zero, and as a result, no work is being done by air resistance on the raindrop. So the correct answer is indeed 0.
For your second question, if the normal force is always vertical, then it would only have a component in the j direction (j-hat) and no component in the i direction (i-hat). So when expressing the normal force in terms of unit vectors i or j, you would only report -fsin(theta) (j-hat) as the only normal force acting on the skier.