Constant velocity and work done

[Samia qureshi]

if a body is moving with constant velocity. Its work done will be? in my point of view Work done is change in energy. Constant velocity means no change in energy. So work done is zero am i right?

 

[Mr-R]

Yes, in the absence of fields the work done is equal to the difference of Kinetic energies at two different points: $$W=\Delta KE.$$ Since the velocity is constant, then $$\Delta KE=0.$$

 

[Samia qureshi]

Yes, in the absence of fields the work done is equal to the difference of Kinetic energies at two different points: $$W=\Delta KE.$$ Since the velocity is constant, then $$\Delta KE=0.$$

Thank you

 

[CWatters]

Perhaps read up on the Work Energy Theorem. This says that the work done by all forces acting on a particle equals the change in the kinetic energy of the particle. In some cases you have to consider the negative work done by gravity or air resistance.

 

[Mr-R]

I think I should have written in the absence of non-conservative fields :)

 

[Samia qureshi]

Perhaps read up on the Work Energy Theorem. This says that the work done by all forces acting on a particle equals the change in the kinetic energy of the particle. In some cases you have to consider the negative work done by gravity or air resistance.

thank you :)

 

[rude man]

I think I should have written in the absence of non-conservative fields :)

That still leaves out gravity, if the object is moving partially or wholly in a gravitational field. And gravity is conservative!

 

[Mr-R]

That still leaves out gravity, if the object is moving partially or wholly in a gravitational field. And gravity is conservative!

You are absolutely right. The work-energy theorem is true for general forces regardless of them being conservative or not (depends on the resultant force). $$\Sigma W=\Delta KE$$ where $$\Sigma=W_c+W_{nc}$$ Conservative and non conservative, respectively. Is this correct?

 

[rude man]

You are absolutely right. The work-energy theorem is true for general forces regardless of them being conservative or not (depends on the resultant force). $$\Sigma W=\Delta KE$$ where $$\Sigma=W_c+W_{nc}$$ Conservative and non conservative, respectively. Is this correct?

Yes, although more conventionally we say that the work done on a mass equals the gain in its potential plus kinetic energy. You have essentially conflated p.e. into work but I guess that is OK too.