# Energy- Work -Power

English version

## Work

In physics, work is related to the amount of energy transferred to or from a system by a force. It is a scalar-valued quantity with SI units of Joule.

Work can be represented in a number of ways. For the case where a body is moving in a steady direction, the work done by a constant force {\displaystyle F}acting parallel to the displacement {\displaystyle \Delta x} is defined as

{\displaystyle W_{F}=F~\Delta x\,\!.}

When the force is not acting parallel to the body’s direction of movement, the work done is defined as a dot product of the force and the displacement,

{\displaystyle W_{F}={\vec {F}}\cdot \Delta {\vec {x}}=||{\vec {F}}||\cdot ||\Delta {\vec {x}}||\cdot \cos \phi \,\!.}

A few other ways of finding work can be either with the change of {\displaystyle K} for kinetic energy or the change of {\displaystyle P} for potential energy which can be resembled as:

{\displaystyle W=\Delta {K}}
{\displaystyle W=\Delta {P}}

In order to define the work done when the force acting upon the body is not constant, we must use differentials to show the infinitesimal work done by the force over an infinitesimal displacement.

{\displaystyle \mathrm {d} W_{F}={\vec {F}}\cdot \mathrm {d} {\vec {s}}\,\!}

### Example

A wagon displaces by a distance of 2 m while under the influence of an 80 N force directed parallel to the motion. How much work is performed by the force exerted on the wagon?

{\displaystyle W_{F}=F\Delta x=80~{\rm {N}}\cdot 2~{\rm {m}}=160~{\rm {N\cdot m}}=160~{\rm {Joules}}\,\!.}

### Example

Suppose the same displacement of 2 m for the wagon while under the influence of an 80 N force 60o to the axis of the motion. How work is performed by the force exerted on the wagon in this case?

{\displaystyle W_{F}=F\Delta x\cos(60^{\circ })=80~{\rm {N}}\cdot 2~{\rm {m}}\cdot 0.5=80~{\rm {N\cdot m}}=80~{\rm {Joules}}\,\!.}

## Power

Power is defined to be the rate at which work is performed, or the derivative of work over time. The SI unit for power is the watt.

{\displaystyle P={\frac {\mathrm {d} E}{\mathrm {d} t}}={\frac {\mathrm {d} W}{\mathrm {d} t}}}

Average power is the average amount of work done per unit of time. Thus instantaneous power is the limiting power of the average power as Δtapproaches zero.

{\displaystyle P=\lim _{\Delta t\rightarrow 0}{\frac {\Delta W}{\Delta t}}=\lim _{\Delta t\rightarrow 0}P_{\mathrm {avg} }}

When the work is done steadily (constant power), just use P = W/t. That is, the power is the work done divided by the time taken to do it.

Example: A garage hoist steadily lifts a car up 2 meters in 15 seconds. Calculate the power delivered to the car. Use 1000 kg for the mass of the car.

First we need the work done, which requires the force necessary to lift the car against gravity:

F = mg = 1000 x 9.81 = 9810 N.

W = Fd = 9810N x 2m = 19620 Nm = 19620 J.

The power is P = W/t = 19620J / 15s = 1308 J/s = 1308 W.

## Energy

Energy is stored work. It has the same units as work, the Joule (J).

There are many forms of energy:

Spring energy: Work has been done on a spring to compress or stretch it; the spring has the ability to push or pull on another object and do work on it. The force required to stretch a spring is proportional to the distance it is stretched: F = kx where x is the stretch distance and k is a constant characteristic of the spring (big heavy springs have larger k values). The work done in stretching a spring from 0 to x is the integral of dW = Fdx. Since the force function is linear, we can just take the average force of kx/2 and avoid using calculus:

   W = average F x distance = (kx/2)(x) = ½kx²


Assuming 100% efficiency, the energy stored in a stretched spring is the same as the work done in stretching it, so Spring E = ½kx²

Example: How much energy is stored in a spring with k = 2000 N/m that has been stretched 1 cm away from its equilibrium length?

E = ½kx² = ½(2000)(0.01)²  = 0.1 J


Gravitational potential energy: a mass has been lifted to a height; when released it will be pulled down by gravity and can do work on another object as it falls.

Example: Find the energy stored in a tonne of water at the top of a 20 m high hydroelectric dam.

The long way is to use F = mg and then W = Fd to find the work needed to lift the water up.

The short way is to combine the formulas, replacing F with mg and using h (height) in place of d:

Gravitational energy = W = Fd = mgh

Egravity = mgh = (1000 kg)(9.81 m/s²)(20 m) = 196200 kg m²/s² = 1.96 x 105 J
Kinetic energy: A mass is moving and can do work when it hits another object. Ekinetic = ½mΔV2 = ½m(Vf2-Vi2)

Example: A 8kg ball is moving at 5m/s. EK = ½(8 kg)(5 m/s)2 = 200 J.

Electrical energy: Electrons can flow out of a battery or capacitor and do work on another electrical component such as a light bulb.