Physics work problem

[ktimekiller]

Ok, I am trying to solve this very simple equation, but I encountered a conflict.

The question is a 1000kg car goes from 0m/s to 10m/s in 3 seconds. What is the work?

I know there are two work formulas I can use here.

The kinetic energy formula of 1/2mv^2

The work formula of Force x distance

If I use force x distance, using average velocity of 5m/s x time to get 15m distance, multiplied by the force of 1000kg, I get 15x10^4

But if I use kinetic energy formula, I get an answer of 5x10^4

When I checked my answers, the Kinetic energy formula yielded the correct answer, but I dont know why the force x distance formula didnt give me the same answer.

I know that both force x distance formula, and the kinetic energy formula is used for problems involving an object going a distance.

How come I cant get the same answer on these 2 different equations?

 

[Avoidist]

A force of 1000kg??

There's your problem. Stick to kinetic energy.

---

Doing by work:

F=ma = 1000*10/3 {a=v/t=10/3}

W=Fs = 10000*15/3=5x10^4.

Conflict resolved.

 

[ktimekiller]

well obviously, I found the answer using kinetic energy, but I want to know WHY the other formula doesnt give the same answer.

 

[Avoidist]

Read it again. You said a force of 1000kg, which is a mass and not a force. Force is always in Newtons. That's why your work calculation isn't working.

 

[ktimekiller]

Ok, considering a force of 10000, and distance of 15m, I still dont get the same answer.



The force x distance is a text book forumla, and on other questions similar to this, involving mass and distance, my answer would come out fine.

 

[Avoidist]

You need to find what the force is by using Newton's 2nd Law: Force=mass x acceleration.

The acceleration is given by a=v/t = 10/3 m/s^2. You have the mass of 1000kg, giving a force of 1000*10/3 = (10^4)/3. Stick that force into the equation for work W=Fs with s=15m and you will get the right answer, as I showed in my edited post.

To reiterate, the force is not 10000.

 

[mindless_moder]

the force you're calculating is the force due to gravity which is perpendicular to the displacement, the force you need to calculate is the force that is going parallel to the displacement vector which is f= m*a.

 

[soulslicer]

wait wait wait, arnt you doing it all wrong. Both ways you are only finding an average of the work done, its like assuming the car travels at 10m/s all the way

the correct way to find accurately the work done, is find Acceleration first, ((10-0)/3)=3.33333..m/s^2
then F=ma, 3.3333..* with 1000 = approximately 3500+ of joules of work done

but frankly im all confused as well. stupid physics.

 

[The Monkey]

I'm so glad I'll never have to do that shit again.

 

[Avoidist]

wait wait wait, arnt you doing it all wrong. Both ways you are only finding an average of the work done, its like assuming the car travels at 10m/s all the way

the correct way to find accurately the work done, is find Acceleration first, ((10-0)/3)=3.33333..m/s^2
then F=ma, 3.3333..* with 1000 = approximately 3500+ of joules of work done

but frankly im all confused as well. stupid physics.

You now have the force of 3333N, W=Fd=3333*15=50000J=5x10^4 J.

Physics is the opposite of stupid, lad.

Edit: Hang on, I'm re-checking that distance calculation, seems a little funny.

Edit #2: Nope, even after integrating d=15m and supports the k.e. calculation.

Edit #3: The proper calculation of the work done by integration of F.dr effectively means that you can take the average and you will get the same result regardless of how the car accelerated over time, which complies with conservation of energy laws seeing as we are assuming it is a conservative system.

F.dr=dr.F=dr.dp/dt=dp.dr/dt=v.dp=vd(mv)=mvdv, Therefore integ{F.dr}=integ{mvdv}=(1/2)mv^2

 

[bbson john]

Use: K.E. = 1/2mv^2 --------------------(1)

Or: W = Fs
= ma*s
=m(v/t)*1/2(at^2)
=m(v/t)*1/2(vt)--------------------------(2)


Same answer is acquired is you sub the variables (m, v, t) into the two formulae.

Since Kinetic Energy formula is so easy, why bother work done?