Work Definition: Force x Distance vs. Force x Displacement

  • Thread starter Thread starter Fizziks_Fan
  • Start date Start date
  • Tags Tags
    Work
AI Thread Summary
The discussion centers on the definition of work in physics, specifically contrasting the dot product of force and displacement with the concept of force multiplied by distance. It clarifies that when lifting and lowering a weight, the net work done is zero due to zero displacement, despite the work done against gravity. The conversation highlights that while lifting requires energy, the work done is technically negative when lowering the weight, as the direction of displacement opposes the force. Additionally, the physical exertion felt by the person is attributed to muscular work, even if the net mechanical work is zero. Overall, the distinction between mechanical work and physical exertion is emphasized.
Fizziks_Fan
Messages
6
Reaction score
0
Hi I always thought work was defined as the dot product of the force and displacement vectors.

W = F parallel to D

However, my physics teacher told me that it's force x DISTANCE.

Does that mean that whenever someone lifts a weight and brings it back down is actually doing work? (displacement is zero...) :rolleyes:
 
Physics news on Phys.org
I've always known work as W = \vec {F} \cdot \vec {D} or W = |F||D|cos\theta theta being the angle between the force and displacement vectors.

More info on work can be found http://en.wikipedia.org/wiki/Mechanical_work
 
The person does no net work when moving the block up and then back down. As you noted, the displacement is 0. You could also argue that it takes say 100 joules of work to move it up, and -100 joules of work to bring it down. However you slice it, it all adds up to nothing.
 
Distance that you are referring to is the distance moved the DIRECTION of the force. The person did do work: against gravity. When he let's go, it is actually gravity that does work, not him.
 
QuantumCrash said:
Distance that you are referring to is the distance moved the DIRECTION of the force. The person did do work: against gravity. When he let's go, it is actually gravity that does work, not him.
]But if he doesn't let go, and slowly lowers the weight to the floor, he still does work, but the net work he does in both directions is zip.
He might be tired and claim he did a lot of work, but he really didn't do any. Depends on what you mean by "work".
 
Well, if you put it that way yes. His total work against is technically negative since the direction of movement when the net work is 0. Since while he lowers down the weight, the direction of the displacement is opposite that of the force. Hence, work done is actually negative.

Note: The man feels tired because his muscles are actually doing work by continously stretching and compressing. I won't delve further since we are already touching biology.
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top