Confusion about the equation KE=1/2mv^2

[Sangari Indiraj]

Summary: at what circumstances the kinetic energy equation can be applied?

My teacher says that, when a constant force is applied and the object moves by d meters, then the work done can be expressed 1/2mv2 where v is the final velocity.
But, what happens if the object moves in a constant velocity in a frictionless surface? can we apply the kinetic energy equation at this time?
In what circumstances this equation is applicable?

 

[anorlunda]

Summary: at what circumstances the kinetic energy equation can be applied?

But, what happens if the object moves in a constant velocity in a frictionless surface? c

In the frictionless case, there is no force and no energy required to move at constant velocity.

But accelerating from zero to constant velocity v requires force, and the energy needed will be 1/2 mv².

But with friction it does take force to move at constant velocity. That uses energy, but the energy in that case goes to heat at the surfaces experiencing friction, not in kinetic energy.

So when working with Newton's Laws, you must be careful to correctly describe the system. An incorrect description leads you to an incorrect conclusion.

 

[PeroK]

In what circumstances this equation is applicable?

In all cases. The KE of a particle with speed $v$ is always $\frac 1 2 mv^2$. This may be a constant speed or an instantaneous speed. Note that speed is the magnitude of velocity, which is in general a vector.

How the particle attained the speed $v$ is another question.

 

[jack action]

My teacher says that, when a constant force is applied and the object moves by d meters, then the work done can be expressed 1/2mv2 where v is the final velocity.

That is not entirely true. The work done is:
$$W = \frac{1}{2}m(v_f^2 - v_i^2)$$
From the basic definition of work:
$$dE= Fdx$$
Assuming that the body is accelerating:
$$dE = madx$$
$$dE = m\left(\frac{v}{dt}\right)dx$$
$$dE = mv\left(\frac{dx}{dt}\right)$$
$$dE = mvdv$$
$$\int_{E_f}^{E_i} dE = \int_{v_i}^{v_f} mvdv$$
$$E_f - E_i = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$$
$$W = \frac{1}{2}m(v_f^2 - v_i^2)$$
So the work done is really the energy difference between two states.

But, what happens if the object moves in a constant velocity in a frictionless surface?

If the velocity is constant, you can see from the equation that there is no work done ... anyway related to acceleration. If there was friction $F_f$ AND a constant velocity - i.e. there is a displacement $\Delta x$ - then there would be work done that amount to $F_f\Delta x$.

In what circumstances this equation is applicable?

When there is an acceleration. Used in the simple form you presented ($\frac{1}{2}mv^2$), it represents the potential kinetic energy that could be extracted by decelerating the object to a stop.

 

[Sangari Indiraj]

That is not entirely true. The work done is:
$$W = \frac{1}{2}m(v_f^2 - v_i^2)$$
From the basic definition of work:
$$dE= Fdx$$
Assuming that the body is accelerating:
$$dE = madx$$
$$dE = m\left(\frac{v}{dt}\right)dx$$
$$dE = mv\left(\frac{dx}{dt}\right)$$
$$dE = mvdv$$
$$\int_{E_f}^{E_i} dE = \int_{v_i}^{v_f} mvdv$$
$$E_f - E_i = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$$
$$W = \frac{1}{2}m(v_f^2 - v_i^2)$$
So the work done is really the energy difference between two states.

If the velocity is constant, you can see from the equation that there is no work done ... anyway related to acceleration. If there was friction $F_f$ AND a constant velocity - i.e. there is a displacement $\Delta x$ - then there would be work done that amount to $F_f\Delta x$.

When there is an acceleration. Used in the simple form you presented ($\frac{1}{2}mv^2$), it represents the potential kinetic energy that could be extracted by decelerating the object to a stop.

So, when an object moves in a constant velocity, in a frictional surface, the work we do is the work done against friction instead of change in kinetic energy?

 

[anorlunda]

So, when an object moves in a constant velocity, in a frictional surface, the work we do is the work done against friction instead of change in kinetic energy?

Yes.

 

[jack action]

So, when an object moves in a constant velocity, in a frictional surface, the work we do is the work done against friction instead of change in kinetic energy?

Yes. The definition of work is a force that moves. It doesn't specify the origin of the force.

If an object accelerates or decelerates (i.e. change in kinetic energy), there must be a force implicated somehow. It's really that force that does the work (not the $m\vec{a}$ vector which is only reacting to the force).

If two opposing but equal forces act on a moving body (i.e. no change in kinetic energy), each does work but cancels each other out (work can be negative), so no work done on the body as a whole.

 

[anorlunda]

In what circumstances this equation is applicable?

You should be confident that energy is always conserved, and that in mechanics work = force * distance. But it is easy to get confused about where the energy goes, or about how much work is done by which object. If the theories appear to be wrong, then you made a mistake somewhere.