## anonymous 4 years ago NEED HELP IN PHYSICS! - mechanical energy. please

1. anonymous

A 200 kg car is traveling at 5 meters per second. what is its total mechanical energy relative to the ground?

2. anonymous

Mechanical energy is kinetic energy (KE) and potential energy (PE). Since the car is on the ground $$PE = 0$$. Therefore, $$ME = KE$$ and $$KE = (1/2) \cdot m \cdot v^2$$

3. anonymous

alright thanks & what about this one. A constant force of 500 N is applied to 50m to the same car in the above question by its engine. What is the new mechanical energy of the car, after the force was applied for 50 m?

4. anonymous

Let's remember that$W = \Delta U$where $$\Delta U$$ is the change in mechanical energy and can be both kinetic or potential. Let's also recall that$W = F \cdot x$Therefore, $F \cdot x = KE_f - KE_i$since the height of the car presumably doesn't change. $$KE_f$$ is the kinetic energy of the car after the force has been applied and $$KE_i$$ is the kinetic energy before.

5. anonymous

What is its new kinetic energy? what is its new speed? what is the increase in its speed? and how much power was applied by the engine of the car if we took 5s to cover the above distance?

6. anonymous

so then the new mechanical energy would be 25000 J ?

7. anonymous

New kinetic energy if $$KE_f$$ from above equation. The new speed comes from the new kinetic energy. I'll assume you know you definition of kinetic energy. The increase in it's speed is the difference between the old and new speeds. Power is defined as work over time. Therefore, $P = {W \over t} = {F \cdot x \over t}$

8. anonymous

That is the amount of the energy increases. Remember the car has energy before we increase, so this initial energy needs to be added to the increase. This can be seen in the equation I gave.

9. anonymous

so for the new mechanical energy of the car, after the force was applied is the same? doesnt change?

10. anonymous

No. New mechanical energy is$KE_{new} = {1 \over 2} 200 \cdot (5)^2 + 500 \cdot 50$ This is the initial kinetic energy PLUS the added kinetic energy.

11. anonymous

and so the new kinetic energy would be the answer to the new mechanical energy.?

12. anonymous

Yes. They are the same in this case because the potential energy does not change.

13. anonymous

and for the new speed what equation would i use?

14. anonymous

$KE = {1 \over 2} m \cdot v^2$

15. anonymous

so it would be 27500 = .5 *2000*v^2 ?

16. anonymous

Yes. Then solve for v.

17. anonymous

and for the increase of speed i just do whatever i got minus the beginning speed?

18. anonymous

Indeed.

19. anonymous

for the power applied which f *x do i use? the new one?

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