Calculating Work in Terms of Final Velocity and Time?

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SUMMARY

The discussion focuses on calculating the work done on a body of mass m that accelerates uniformly from rest to a final velocity \( v_{f} \) in a time \( t_{f} \). The derived formula for work as a function of time is \( \frac{1}{2}m\frac{v_{f}^2}{t_{f}^2} t^2 \). Key equations utilized include \( W=\int F*dx \), \( V_{f}=at \), and \( F=m\frac{V}{t} \). The solution emphasizes that for constant force, work is simply the product of force and distance, while varying forces require integration.

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Homework Statement


A body of mass m accelerates uniformly from rest to a speed v_{f}
in time t_{f}
Show that the work done on the body as a function of time, in terms of v_{f},t_{f} is:
\frac{1}{2}m\frac{v_{f}^2}{t_{f}^2} t^2

Homework Equations


1)W=\int F*dx
2)V_{f}=at
3)F=m\frac{V}{t}

The Attempt at a Solution


Well I know I would start out integrating equation 1 with equation 3.
Then what?
 
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Since the acceleration is uniform (i.e. constant) So is the force and thus the integral is not required. You're on the right lines with equation 3, but you need to consider the distance traveled by the object aswell. Think of the uniform acceleration equations.
 
Ah of course I get it now:
r=\frac{1}{2}at^2
W=F*r \longrightarrow \frac{1}{2}ma^2t^2 \longrightarrow \frac{1}{2}m\frac{v^2}{t_{f}^2}t^2
 
So the thing to remember from this about work is that if the force is constant then its simply the force multiplied by the distance (or in vector terms the dot product) and if the force varies you have to integrate.
 

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