Calculating Work in Terms of Final Velocity and Time?

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The discussion focuses on deriving the work done on a body accelerating uniformly from rest to a final velocity, v_f, over a time period, t_f. The key equations used include the work-energy principle and the relationship between force, mass, and acceleration. The solution involves integrating the force with respect to distance and recognizing that with uniform acceleration, the distance can be expressed as r = (1/2)at^2. The final expression for work done is derived as W = (1/2)m(v_f^2/t_f^2)t^2. Understanding the relationship between constant force and distance is crucial for calculating work in this context.
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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.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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