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## Class 11 Physics (India) - Hindi

### Course: Class 11 Physics (India) - Hindi > Unit 4

Lesson 8: Power (Hindi)# Power review

Review the key concepts and equations for power, including the difference between power and energy.

## Key terms

Term (symbol) | Meaning | |
---|---|---|

Power ( | Rate at which work is done (or energy is transferred). SI units of | |

Watt ( | Power equivalent to transferring |

## Equations

Equation | Symbols | Meaning in words |
---|---|---|

Power is the change in energy over the change in time. | ||

Power is the work done over the change in time. |

## Common mistakes and misconceptions

**Some people confuse energy with power.**Energy is a measurement of work done, while power is how energy changes with time.**Sometimes people forget that we can measure power by the work done by a force.**We can substitute the work done by a force instead of change in energy into the equation$W=F\mathrm{\Delta}x\mathrm{cos}\theta $

If the force is along the direction of motion, then $\mathrm{cos}\theta =1$ and the equation can be re-written

where $F$ is a constant force, and a change in distance over time is average velocity, $\overline{v}$ .

## Learn more

For deeper explanations, see our video introducing power.

To check your understanding and work toward mastering these concepts, check out the exercise on relating power and energy.

## Want to join the conversation?

- A car weight 2500 N operating at the rate of 130 kW develops a maximum speed of 31 m/s on a level, horizontal road. Assuming that the external force (due to friction and air resistance ) remains constant,What is the car’s maximum speed on an incline of 1 in 20 ( i.e if θ is the angle of the incline with the horizontal, sin θ = 1/20 ) ?(2 votes)
- I can let you know the strategy for solving this problem.

First you need to find the force applied to the car when it is traveling on a level road. Use the power formula to find this. P=Fv From this formula you can see that using 130000W and 31m/s we can find the force to be 4193N. Since the car was on a level road and traveling at a constant velocity then that means the friction force by air resistance and drag must also be 4193N.

We now move on to the second phase of the problem. The problem said to assume the resistance force is equal. So in this case the resistance force as it travels up the incline is also 4193N. At this point you need to draw a free-body diagram(FBD) of the forces involved acting on the car as it travels up the incline. I can not do that for you here so you will need to do that yourself. After analyzing the FBD you should discover that the Force acting on the car from the ground is

`F = 4193 + 2500*sin(`

*theta*) = 4193 + 2500(1/20) = 4318N

You now have the force applied to the car by the ground. This is just the third law pair of the force the car applied to the ground. So the car output a force of 4318N on the incline. But you also know the power output of the car to be 130000W. So using the power formula again we can discover the velocity of the car on the incline.`v = P/F = 130000W/4318N =`

**30.1m/s**

So the maximum speed of the car on the incline is**30.1m/s**. This is only slightly slower than the original 31m/s on the level road and that is reasonable since the incline is very small, 2.87 degrees.(11 votes)

- How can I create an equation for the time from this formula(1 vote)
- You can create an equation for the time from the power formula by using some algebraic manipulation.
**P = ▲E /▲T**(Given)**P · ▲T = ▲E**(Multiplying by ▲**T**on both sides)

▲**T = P /▲E**(Dividing by**P**on both sides)

We have now isolated ▲**T**(**time**) so that it can be written in terms of**P**(**power**) and ▲**E**(**energy**).(2 votes)

- From the last part where they proved P=Fv, shouldn't the 'd' from Fdcos(θ) be distance instead of displacement (Δx)?

Then P will be force times speed instead of force times velocity.(1 vote) - whats the difference between potential energy and just potential(1 vote)