# Relationship between net force and kinetic energy

Considering work-energy principle we know net-work done on a system is equal to the change in kinetic energy of the system. The net-force causes net-work. So can we also state net-force causes acceleration in the direction of the net force and so also causes a change in kinetic energy of the system?

(Asking this because I read a derivation of Bernoulli's equation, which can be taken as an equation of conservation of mechanical energy, from Newton's second law. Newton's second law used for the derivation of momentum equation made sense but second law being just about force and change in momentum didn't quite intuitively get how it was used to relate to conservation of energy. I got the math behind it which is straightforward just didn't get a sense for it.)

• Net force could cause a deceleration, which would decrease KE. – Kyle Kanos Nov 26 '17 at 17:02
• @KyleKanos Yup!! I meant change in kinetic energy...have edited my question. – GRANZER Nov 26 '17 at 18:32

More completely, with $K$ the kinetic energy, $U$ the potential energy and $W$ the work done, then:

$$W=\Delta K+\Delta U$$

If there's no change in $U$ ($\Delta U=0$), then obviously:

$$W=\Delta K$$

Newton's Second Law tells us, with $F$ the force acting on a body of mass $m$, for simplicity's sake we'll consider $F=\text{constant}$:

$$F=ma$$

Let's do this in one dimension, for simplicity's sake, i.e. $x$:

The force causes a displacement $dx$ and performs work $dW$ on the mass $m$:

$$dW=Fdx=madx$$

A little trick:

$$a=\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=v\frac{dv}{dx}$$

Insert into $dW$:

$$\implies dW=mvdv$$

So that:

$$\int_0^WdW=\int_{v_1}^{v_2}mvdv$$

$$\boxed{W=\frac12 mv_2^2-\frac12 mv_1^2=\Delta K}$$

As written elsewhere, $W$ can take on any value, positive or negative (or zero). A braking force for instance will act in the opposite sense of the velocity vector and cause deceleration, so that $v_2<v_1$ and $W<0$, $\Delta K<0$. So mind your signs!

• $W=12mv22−12mv21=ΔK$ this is always true when talking about net-work ryt...because net-work is always only equal to change in kinetic energy. – GRANZER Nov 26 '17 at 18:47

Of course we can, that is Newton's second law! $\vec{F}_{net}=m\vec{a}$

The one caveat is, as Kyle mentioned, acceleration ($\vec{a}$) is a vector quantity so the kinetic energy could increase or decrease (or even stay the same!) depending on the direction of the net force with respect to the direction of the velocity of the particle. For the kinetic energy to stay the same, the force would have to be applied perpendicular to the velocity. Think (circular) orbits!

So can we also state net-force causes acceleration in the direction of the net force and so also causes an increase in kinetic energy of the system?

$$KE = \frac{1}{2}mv^2 = \frac{1}{2}m\vec v\cdot\vec v$$

$$\frac{d}{dt}KE = \frac{1}{2}m\frac{d\vec v}{dt}\cdot\vec v\, + \frac{1}{2}m\vec v \cdot \frac{d\vec v}{dt} = m\vec a \cdot \vec v = \vec F \cdot \vec v = Fv\cos \theta$$

where $-\pi \lt \theta \le \pi$ is the angle between the force and velocity vectors.

Thus, for non-zero net force, the change in kinetic energy in time $dt$ can be positive $(|\theta| \lt \pi/2)$, zero $(|\theta| = \pi/2)$, or negative $(\pi/2\lt|\theta|\le\pi)$

• Thank you. So it is right to say net force does causes a change in kinetic energy and t is equal to net -work. – GRANZER Nov 26 '17 at 19:36
• @GRANZER, if the net force is always perpendicular to the velocity, e.g., uniform circular motion, the KE does not change. – Alfred Centauri Nov 26 '17 at 21:21

## protected by Qmechanic♦Nov 26 '17 at 18:55

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