You drop a box, mass m = 3.0 kg, from 10 meters above the ground. The initial sp
ID: 1587802 • Letter: Y
Question
You drop a box, mass m = 3.0 kg, from 10 meters above the ground. The initial speed is zero. Now because of air drag, the box reaches a terminal speed of 6.1 m/s, when it reaches 5 meters above the ground. So.....the speed is 6.1m/s from 5 meters down to zero meters, just before it hits the ground. Obtain the work done by gravity, from 10 meters to 5 meters. Obtain the change in KE from 10 meters to 5 meters. Using the work-kinetic energy theorem, obtain the work done by air drag from 10 meters to 5 meters. Obtain the work done by gravity, from 5 meters to 0 meters. Obtain the change in KE from 5 meters to 0 meters. (Note: The answer is ZERO.) Using the work-kinetic energy theorem, obtain the work done by air drag from 5 meters to 0 meters. Explain the difference between your answers to (c) and (f) above. What's going on?Explanation / Answer
a) Workdone by gravity(from 10 to 5m) = m*g*h
= 3*9.8*5
= 147 J
b) change in KE(from 10 to 5m) = 0.5*m*(vf^2 - vi^2)
= 0.5*3*(6.1^2 - 0^2)
= 55.815 J
c) Net Workdone = change in kinetic energy
W_gravity + W_airdrag = 55.815
147 + W_airdrag = 55.815
W_airdrag = 55.815 - 147
W_airdrag = -91.185 J
d) Workdone by gravity(from 5 to 0m) = m*g*h
= 3*9.8*5
= 147 J
e) change in KE(from 10 to 5m) = 0 (since the body moves with constant speed)
f) Net Workdone = change in kinetic energy
W_gravity + W_airdrag = 0
147 + W_airdrag = 0
W_airdrag = -147 J
g) As the body falls down air resistve forces increases, whenever body gets terminal speed the air resistive force is exactly equal to gravity.
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