Work Question - Is something missing?

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To calculate the work done in accelerating a 2.0 kg object from 2.0 m/s to 3.0 m/s, the necessary information includes mass, initial velocity, and final velocity. While the initial equation provided was correct, it lacked the necessary relation involving kinetic energy. The correct approach is to use the change in kinetic energy formula, which is (1/2) m (vf^2 - vi^2). This leads to the conclusion that the work done is 5 Joules. Understanding this relation simplifies the calculation and avoids unnecessary derivations in future problems.
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Homework Statement


Calculate the work done accelerating a 2.0 kg object from 2.0 m/s to 3.0 m/s.

mass=2.0 kg
Initial Velocity= 2.0 m/s
Final Velocity= 3.0 m/s


Homework Equations


Work=applied force x displacement



The Attempt at a Solution


Correct me, if I am wrong, but don't I need a bit more information to answer this question? This was all the information given to me.
 
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Hi Sabellic,

Sabellic said:

Homework Statement


Calculate the work done accelerating a 2.0 kg object from 2.0 m/s to 3.0 m/s.

mass=2.0 kg
Initial Velocity= 2.0 m/s
Final Velocity= 3.0 m/s


Homework Equations


Work=applied force x displacement



The Attempt at a Solution


Correct me, if I am wrong, but don't I need a bit more information to answer this question? This was all the information given to me.

No, that's all you need. The equation you listed is true; however, there is another equation that involves work and velocity. What is it?
 
Hmmmm. I don't know. The only thing that I can conjure up is that the work can be equivalent to (mass/acceleration) * (final velocity^2 - initial velocity^2)/2 * a^2

But that doesn't work because i don't have the acceleration.
 
Sabellic said:
Hmmmm. I don't know. The only thing that I can conjure up is that the work can be equivalent to (mass/acceleration) * (final velocity^2 - initial velocity^2)/2 * a^2

But that doesn't work because i don't have the acceleration.

That's so close! But it needs to be:

(mass/acceleration) * (final velocity^2 - initial velocity^2)/2 * a

and so the accelerations will cancel.

Then notice that what you have left is:

(1/2) m vf^2 - (1/2) m vi^2

Does that look more familiar? What are those terms? Once you have that, you will have the important equation that you need to know.
 
Oh God you're right. I wrote (mass/acceleration) instead of mass * acceleration. So yes, they would cancel out. Thanks a lot for this. So it will be 5 Joules, I think.
 
That's good, but you don't want to have to derive that important formula every time. The way to think about it is

total work = change in kinetic energy

(where total work includes all conservative and non-conservative work).

Since kinetic energy is (1/2) m v^2, you could write down the relation you need right away.
 
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