In preparing for class today I was reviewing some problems in
Halliday, Resnick and Walker (5th ed) - Chapter 7. I believe the answer
to problem 15E on page 150 is incorrect.
The problem states that a 12 m long firehose is uncoiled by
pulling the nozzle end horizontally along a frictionless surface at the
steady speed of 2.3 m/s. The mass/length of the hose is 0.25kg/m. How
much work has been done on the hose by the applied force when the entire
hose is moving?
The solution manual applies the "Work-Energy" theorem and
calculates 0.5mv^2 for the entire hose in motion to get the answer 7.9 J.
However, I believe that this is an incorrect application of the W-E
theorem because the mass being moved by the applied force is not constant.
i.e. from Newton's second law F = d(mv)/dt = (dm/dt)v in this case, so the
total work done is the integral of Fdx, which with a little manipulation I
calculate to be rho*v^2*L where rho is the mass per unit length of the
hose and L is the total length of the hose. This gives an answer twice as
big.
Since I'm not a "morning person", have I blundered?
Mark
Dr. Mark H. Shapiro
Physics Department
California State University, Fullerton
P.O. Box 6866
Fullerton, California 92834-6866