Check out some similar questions!

johnsonscience · Oct 5, 2010, 05:09 AM

An 80kg man is standing on a 140kg pontoon boat. The system is motionless. The man dives off. His dive is perfectly horizontal (no PE change). The man goes 7 m/s to the left and the pontoon goes 4 m/s to the right. How much internal energy did the man convert to KE? The KE to the left would be 1960J and the KE to the right would be 1140J. It seems to me that the 820J difference would be dissipated energy due to friction between the boat and the water. So the total overall kinetic energy would be 1960J x 2 = 3920. Is this amount all from the internal energy of the diver? Or is it half of that because the pontoon is experiencing a reaction force? Also, if this could be counted as an explosive collision how would the formula for kinetic energy read? I thought 0 + 0 ----> 1960J + 820J + 1140J but I am not sure this is correct

ebaines · Oct 6, 2010, 05:46 AM

You seem to think that the energy "to the left" must equal the energy "to the right," but this is not true. Kinetic energy is a scalar, not a vector, so it has no direction. The total KE is simply is the KE of the man (1960 J) plus the KE of the boat (1120 J), or 3080 J total.

BTW, check your math for the KE of the boat - I get
$<br /> KE_{Boat}\ = \ \frac 1 2 \times 140\ Kg \times (4 m/s)^2 = 1120\ J <br />$

johnsonscience · Oct 7, 2010, 05:16 AM

Okay! No, I meant to type 1120J!
Is there any dissipated energy?

ebaines · Oct 7, 2010, 05:38 AM

You asked about "dissipated energy." It seems pretty clear that in the real world the boat and the man will slow down due to friction with the water molecules which creates an opoosing force to their movement. Some of their kinetic energy will be transferred to the water (ceating waves and eddies) and some will turn into heat. However, at the instant that the boat is moving at 4 m/s and the man is movig at 7 m/s, their total KE is as shown in the earlier post.