A 100g ball is dropped from a height of 25m and reaches the ground with a speed of 10m/s. What fraction of the original energy is lost to air resistence? (The answer is 80% but how do calculate it?)
-
Conservation of energy
Kinetic Energy + Potential Energy = Total Energy of the system
Since the ball is DROPPED, there is no Kinetic Energy at a height of 25m
therefore, the Total Energy of the system is just the Potential Energy at 25m = m*g*h =
(.1)(9.81)(25) = 24.525 J
This is the amount of energy that SHOULD BE at the bottom of the balls fall, but
Total Energy when the ball hits the ground is just Kinetic Energy = 1/2(m*v^2) =
1/2(.1)(100) = 5 J
The Total Energy lost is 24.525 - 5 = 19.525 J
19.525 / 24.525 = .796
Total % Energy lost = (.796)*(100) = 80%
Damn you air!
Kinetic Energy + Potential Energy = Total Energy of the system
Since the ball is DROPPED, there is no Kinetic Energy at a height of 25m
therefore, the Total Energy of the system is just the Potential Energy at 25m = m*g*h =
(.1)(9.81)(25) = 24.525 J
This is the amount of energy that SHOULD BE at the bottom of the balls fall, but
Total Energy when the ball hits the ground is just Kinetic Energy = 1/2(m*v^2) =
1/2(.1)(100) = 5 J
The Total Energy lost is 24.525 - 5 = 19.525 J
19.525 / 24.525 = .796
Total % Energy lost = (.796)*(100) = 80%
Damn you air!
-
PE = KE that is assuming no loss to air friction
So the actual KE / KE above equals the % KE remaining
So the actual KE / KE above equals the % KE remaining