How Does Doubling Power Output Affect a Kayaker's Speed?

[harmeet_angel]

Homework Statement

If a kayaker doubles her power output, and the resistive force due to the water remains same, by what factor does her speed changes(provided that her initial speed was constant, and is v0)?

Homework Equations

Power = Fd/t

The Attempt at a Solution

So, first I found the resistive force, using initial power (Pi) and initial speed (v0)
Pi=F(v0)
F=air resistance since the net force is0
Therefore I got the air resistance to be:
F=(Pi)/(v0)

And if she doubles her power, that means her applied force would be more than air resistance, and thus there would be net acceleration, and I can find new v using time or distance. But neither of them is given in the question.. so I couldn't find another way to solve this problem.

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and for the questions that attached along with this post, I am wondering about the second question(#68) it consists of two strings,
if it means that word needed it
0.5x^2(k2+k1)-mgx?
(work required to stretch the first string, and then the second string, and as gravity helps)
http://ca.geocities.com/cheemaharmeet/lastscan.jpg
and for those two questions do i need to subtract by mgx?
cuz they say work required, so if that means we also have to exclude gravitational work?

 

[Nate810]

For the second question, you need to subtract out the work done by gravity. The work required is the work needed to stretch the two strings, so you need to subtract out any work done by gravity as that would not be part of the required work.

 

[m2003]

For the first question, you are correct in finding the resistive force due to air. However, since the kayaker is doubling her power output, the net force will not be 0. The new net force will be the difference between the applied force (doubled power) and the resistive force. Using this net force, you can use the equation F = ma to find the acceleration of the kayaker. Then, using the initial speed and the new acceleration, you can use the equation v = u + at to find the new speed.

For the second question, the work required to stretch the first and second strings would be the sum of the work required to stretch each individual string. So, the expression would be 0.5x^2(k2+k1)-mgx. The gravitational work should not be excluded since it is still a force acting on the system and would contribute to the overall work required.