Kinetic Energy

Kinetic energy is the energy of motion.

We would like to express the kinetic energy of an object in terms of properties like mass $m$ and velocity, $v$.

Let's start by accelerating from rest by a constant force $F$ acting over a distance $d$.

The work $W$ that we did on the object is $W=Fd$.

The Work-Energy Theorem tells us that the work we did is equal to the change in kinetic energy.

Therefore the kinetic energy which we call $E_K$, $E_K=W=Fd$.

But how can we relate this to the motion of the object ? It would be nice to have an expression for the kinetic energy purely from the parameters of the object itself.

We use Newton's Second Law! This says that $F=ma$, where $m$ is the mass of the object and $a$ is the acceleration of the object. Note that when $F$ and $m$ are constant, $a$ will be constant.

Using this law to substitute for the force, $E_K=mad$, although we are certainly not mad with this result.

When we started out, we wanted to figure out the kinetic energy in terms of the mass of the object and its speed. Now the constant force leads to a constant acceleration, so let's try to use our familiar friends, the kinematics equations to help!

Technically, there was no wrong answer to the above question. We could have used any combination of the formulas to get the answer. However, the nicest answer comes from the equation $v_f^2-v_i^2=2a\Delta x$.

We already know some of these values! Since we are accelerating the object from rest, $v_i=0$. Also, $\Delta x=d$, since that is the distance the object moves!

Thus, letting $v_f=v$, we have the following: $$v^2=2ad$$ Plugging this into our equation, $E_K=mad$, we have: $$E_K=\frac{1}{2}mv^2$$ So there you have it! The kinetic energy of an object is one half of the product of the object's mass and the square of the object's velocity.

You can check that the equation is dimensionally correct. The units on both sides are [kg] [M] /[s$^2$].

Now, we derived this formula for an object accelerated from rest with a constant force, but this formula is true regardless of how an object is accelerates. I'd you know calculus, try repeating this derivation for an object accelerated by a time dependant force $F(t)$.

Kinetic energy

Kinetic energy is the energy of a body which is produced by its motion. How can we calculate the kinetic energy of an object?

We know that the word- 'Energy' means to say the ability of working.

Assume that you have an object.

You are applying F=10N of force restlessly from position A to position B. The displacement between the position A to B is equal to S.

So, how much work will you have done on the object, if you want to displace it from position A to position B? or how much energy will you have spent to move it from position A to B? This is so easy to say. The magnitude of your given energy on the object or the work which has been done by you is $$W = Fs\cos C = FS\cos0 = FS$$ (as the angle between force and displacement is 0).

According to the conservation of energy, we know that energy can not be created or destroyed. So, where has the your given energy gone? Actually, the energy which is applied by you on the object to move has changed into the kinetic energy of the object.

** How to measure the magnitude of Kinetic Energy? **

W= E = maS. Let's find the value of 'a.S' Assume that we have started with 0 initial velocity. So,

then,

So, kinetic energy of a body depends on its mass and the square of its final velocity when the initial velocity is 0. Kinetic energy -

Can you calculate the kinetic energy of an object if it has 'u' initial velocity? This problem is for you to solve.

In this post we'll do some easy Energy calculations, along with momentum.

For more on momentum, look at these pages:

What is Momentum?

Conservation of Momentum

In fact, the coolest thing about momentum is that it is always conserved, even through collisions!

With this, let's start with our situation.

A ball of mass $4$ kilograms is rolling at a velocity of $6 m/s$. It is going straight towards another ball of mass $2$ kilograms. The balls collide, and afterwards, the ball of mass $2$ kilograms moves twice as fast as the ball of $4$ kilograms.

How you do this is just $4 kg \times 6 m/s = 24 \frac{kg \cdot m}{s}$.

Here's how we do it: let the velocity of the ball weighing $4$ kilograms be $v$, and therefore, the velocity of the smaller ball is $2v$. Therefore, by our momentum equation, the total momentum is $v \cdot 4 kg + 2v \cdot 2 kg = 8v kg$. Therefore, by conservation of momentum, $8 kg \cdot v = 24 \frac{kg \cdot m}{s} \implies v = 3 m/s$.

So the large ball is moving at $3 m/s$ and the small ball is moving at $6 m/s$.

Now let's move on to the real point of this exercise: Energy!

How you do this is just using our formula: $KE = \frac{1}{2} mv^2 = \frac{1}{2} 4 \cdot 6^2 = 72$. Hmmm, but I left out something here. Units!

How you find the answer is by using the formula for kinetic energy: the units of $\frac{1}{2} mv^2$ are $kg \cdot \left( \frac{m}{s} \right)^2 = \frac{kg \cdot m^2}{s^2}$. To be lazy, we make a new unit called the Joule ($J$ for short), and $J = \frac{kg \cdot m^2}{s^2}$, so it is the unit for energy.

We find this by summing the energies of the 2 balls. Using our formula, this gives $\frac{1}{2} 4 \cdot 3^2 J + \frac{1}{2} \cdot 2 \cdot 6^2 J = (18 + 36) J= 54 J$.

Note that energy decreased! There is no such thing as "conservation of energy". In fact, in every collision, some energy is dissipated, or released into the environment. So the result is every collision is a loss of energy in our system!