- #1
Jezza
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Homework Statement
A body of mass [itex]M[/itex], traveling in a straight horizontal line, is supplied with constant power [itex]P[/itex] and is subjected to a resistance [itex]Mkv^2[/itex], where [itex]v[/itex] is its speed and [itex]k[/itex] is a constant. Show that the speed of the body cannot exceed a certain value [itex]v_m[/itex] and find an expression for [itex]v_m[/itex].
Homework Equations
[itex]P=\frac{dW}{dt}[/itex]
[itex]F=\frac{dv}{dt}[/itex]
Where [itex]W[/itex] is work, [itex]F[/itex] is force, [itex]t[/itex] is time.
The Attempt at a Solution
First, find the force applied to the body by the power:
[itex]P = \frac{dW}{dt} = \frac{d}{dt}\int{Fdx} = \int{F \frac{dx}{dt}} = \int{Fdv} [/itex]
We conclude by the fundamental theorem of calculus:
[itex]F = \frac{dP}{dv}[/itex]
The resultant force on the body is therefore [itex]F_{res} = \frac{dP}{dv} - Mkv^2 = 0[/itex] at [itex]v_m[/itex].
Now we can write [itex]\frac{dP}{dv} = Mkv^2[/itex] which gives:
[itex]P = \int_{v_0}^{v_m}{Mkv^2dv} = \frac{1}{3}Mk (v_m^3 - v_0^3)[/itex]
If it's ok to assume that [itex]v_0 = 0[/itex] (which I don't think it can be - I don't think it should depend on any initial velocity):
[itex]v_m^3 = \frac{3P}{Mk}[/itex].
The answer is given as [itex]v_m^3 = \frac{P}{Mk}[/itex].
What am I doing wrong? Also why am I having to set an initial velocity? Thanks for any help :)