Work Question - Is something missing?

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Homework Help Overview

The problem involves calculating the work done in accelerating a 2.0 kg object from an initial velocity of 2.0 m/s to a final velocity of 3.0 m/s. The context is rooted in the principles of work and energy in physics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the need for additional information, particularly regarding acceleration, to solve the problem. There is exploration of relevant equations and relationships between work and kinetic energy.

Discussion Status

Some participants have offered guidance on relevant equations and clarified the relationship between work and kinetic energy. There is an ongoing exploration of how to derive the necessary formulas without reaching a consensus on the final approach.

Contextual Notes

Participants note the absence of acceleration in the provided information, which is a point of contention in determining how to calculate work. There is also mention of the need to avoid deriving formulas repeatedly in future discussions.

Sabellic
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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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