A 4.62 x 10-5 kg raindrop falls vertically at constant speed under the influence of gravity and air resistance. After the drop has fallen 83.7 m, what is the work done by gravity? Answer in units of milliJoule (a thousandth of a Joule) and use 9.8 m/s2 for g
m g h = change in potential energy = work done by gravity
= 4.62 * 10^-5 * 9.8 * 83.7 Joules
then multiply by 10^3 to get milli Joules
To find the work done by gravity on the raindrop, we can calculate the gravitational potential energy difference between two points: the initial position and the final position. The work done by gravity is equal to the change in potential energy.
The formula for gravitational potential energy is:
PE = m * g * h
Where:
PE is the potential energy,
m is the mass of the object,
g is the acceleration due to gravity,
and h is the height or vertical displacement.
In this case, we are given the mass of the raindrop as 4.62 x 10^(-5) kg, the height as 83.7 m, and the acceleration due to gravity as 9.8 m/s^2.
So, the potential energy at the initial position is:
PE_initial = m * g * h_initial = (4.62 x 10^(-5) kg) * (9.8 m/s^2) * 0 = 0 J
The potential energy at the final position is:
PE_final = m * g * h_final = (4.62 x 10^(-5) kg) * (9.8 m/s^2) * 83.7 m
To find the work done by gravity, we need to subtract the initial potential energy from the final potential energy:
Work = PE_final - PE_initial
= (4.62 x 10^(-5) kg) * (9.8 m/s^2) * 83.7 m
= 3.88 x 10^(-2) J
To convert from Joules to milliJoules, we need to multiply the result by 1000:
Work = 3.88 x 10^(-2) J * 1000
= 38.8 mJ
Therefore, the work done by gravity is 38.8 milliJoules.