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Work is expressed as $W=Fd$, where the $F$ is in Newton, $d$ is in meters and result $W$ is in Joules. For example, if I take $1N$ on earth and lift it $1m$ up in the air I have done $1J$ of work.

Kinetic energy is expressed as $E=\frac{1}{2}mv^2$, where the $m$ is mass of the body in kilograms and $v$ is the speed of the body in $m/s$ and the result is again in Joules. For example, if I have $80kg$ and run at speed of $5m/s$ then I have $2000J$ of kinetic energy.

I am very puzzled because I do not see the relation between newtons and kilograms in these two examples.

I mean, how can you mix up two different things such are mass and force and get the results in same unit such as Joule - Are those two results are the same Joules?

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Mass and force aren't all that different; a force can be considered a mass under acceleration (F = ma), which resolves the units in the given equations. – Ignacio Vazquez-Abrams Jul 14 '12 at 12:48
Why are you ignoring the units of distance and velocity in your analysis? I mean, you are noting that the LHS of each eqution has the same units and then complaining that one of the two quantities on the RHS differs in two cases. – dmckee Jul 14 '12 at 18:07
@dmckee: It's a fair comment. But because the unit analysis is trivial, probably what the OP is really getting at is what does mass have to do with kinetic energy when work is defined in terms of forces and not masses. Basically, a question aimed at very basic comprehension rather than technical difficulties. – Niel de Beaudrap Jul 15 '12 at 23:35
up vote 3 down vote accepted

Short answer. The relation between Newtons and kilograms, with respect to work / kinetic energy, is actually just through Newton's Second Law!

Here we are not mixing up Newtons and kilograms (forces and masses). The mass plays a role in the kinetic energy equation precisely because mass plays a role in how much force it takes to accelerate an object from rest to a speed v.

Consider for instance an object which has a mass of, say, 0.1019 kg. The gravitational force that this body experiences at the surface of the Earth is −1 N (that is, a downward force), if we take g = −9.81 m/s². If we hold this object from a height of one meter, and allow it to drop, gravity performs work on this object, exerting a force of −1 N over a (downward) displacement of −1 m. As a result, just before impacting the ground, the object will have 1 Joule of kinetic energy, as (−1 N) · (−1 m) = +1 J; gravity will have done 1 Joule worth of work on the object, which causes it to move with 1 Joule worth of kinetic energy.

In the examples you give, the correspondence that you're looking for is between mass and the gravitational force that it exerts. When you say that you take "1 Newton" and lift it one meter, the 'Newton' you're referring is one Newton of force exerted downward by an object with some mass — or more to the point, the one Newton which would be the minimum necessary to raise it steadily by opposing gravity.

We can actually show directly how mass comes into the kinetic energy formula, and pinpoint the reason that it's there.

Suppose that we exert a net force F on an object with mass m, over some displacement d in the same direction as the force, where the object starts at rest. The amount of work done on the object will be $$ W \;=\; \mathbf F \cdot \mathbf d \;=\; Fd,$$ taking F to be the magnitude of the force F, and d to be the length of the displacement d. Because we're doing net work on the object starting from rest, this work will go directly towards the kinertic energy of the object. The acceleration of the object under this force is $$ \mathbf a \;=\; \mathbf F / m . $$ If t is the time that it takes for the object to be moved by a displacement of d, we have $$ \mathbf d \;=\; \tfrac{1}{2}\mathbf a \, t^2 \,;\qquad\implies\qquad t \;=\; \sqrt{2d/a\;} \;=\; \sqrt{2dm/F\;}.$$ taking the magnitude a = || a ||. The speed that the object is travelling after that amount of time is just $$\begin{align*} v \;=\; a t \;&=\; \bigl(F/m\bigr) \sqrt{2dm/F\;} \\[1ex]&= \sqrt{2Fd/m\;}, \end{align*}$$ applying Newton's Second Law again for the acceleration, and cancelling factors of F and m under the square-root.We may then re-express this equation as $$ v^2 \;=\; 2Fd/m \qquad\implies\qquad \tfrac{1}{2}m\,v^2 = Fd = W, $$ where we just applied the formula for the work done on the object at the end. Because the net work is the same as the kinetic energy in this case, it follows that $K = \tfrac{1}{2}m\,v^2$.

In the derivation above, the only ways we used mass was with Newton's Law, a = F/m. So, the fact that it occurs in the formula for kinetic energy isn't because we are confusing mass with force, but because of the relationship between mass and force.

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They are the same Joules. They might have a different algebraic sign, but they are both basically energy. Look at it this way: You need a certain amount of Energy E to lift a mass: $$ E = m*g *h $$ Let:

m := 1kg

g := 9,81 m/s²

h := 1m

This gives you: $$ E =9,81 \frac{kg*m²}{s²}= 9,81 N*m = 9,81 J $$

Now you'll take the same mass and trow it up with the same Energy: $$ E = 1/2 * m * v² $$ $$ v = \sqrt[2]{\frac{2*E}{m}} $$ $$ v =4.43 \frac{m}{s} $$ A little calculation with newtons laws will give you this formula for the maximum hight: $$ h = \frac{v²}{2*h} = 1 m $$ So you get, that the required energy doesn't depend on whether you lift something or throw it.

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So, basically when you throw body up using 9.81J of work it will have have starting speed of 4.43m/s and reach 1m up? – Dusan Jul 14 '12 at 13:12
It depends on the mass. For 1 kg: yes – Stein Jul 14 '12 at 13:38

In words, KE is the energy stored from work done by a force (through a distance) on the particle.

Work is essentially the measure of energy converted from one form to another. For example, a particle in a potential experiences a force, accelerates and gains KE while losing PE.

The PE lost is stored as the KE of the particle. This conversion of PE to KE is work.

The work done by the potential on the particle is precisely the amount of PE converted to KE. The power is the rate at which energy is converted, the time rate of change of work.

See Work-Energy theorem.

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