anonymous
  • anonymous
NEED HELP IN PHYSICS! - mechanical energy. please
Physics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
A 200 kg car is traveling at 5 meters per second. what is its total mechanical energy relative to the ground?
anonymous
  • anonymous
Mechanical energy is kinetic energy (KE) and potential energy (PE). Since the car is on the ground \(PE = 0\). Therefore, \(ME = KE\) and \(KE = (1/2) \cdot m \cdot v^2\)
anonymous
  • 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?

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anonymous
  • 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.
anonymous
  • 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?
anonymous
  • anonymous
so then the new mechanical energy would be 25000 J ?
anonymous
  • 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, \[P = {W \over t} = {F \cdot x \over t}\]
anonymous
  • 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.
anonymous
  • anonymous
so for the new mechanical energy of the car, after the force was applied is the same? doesnt change?
anonymous
  • 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.
anonymous
  • anonymous
and so the new kinetic energy would be the answer to the new mechanical energy.?
anonymous
  • anonymous
Yes. They are the same in this case because the potential energy does not change.
anonymous
  • anonymous
and for the new speed what equation would i use?
anonymous
  • anonymous
\[KE = {1 \over 2} m \cdot v^2\]
anonymous
  • anonymous
so it would be 27500 = .5 *2000*v^2 ?
anonymous
  • anonymous
Yes. Then solve for v.
anonymous
  • anonymous
and for the increase of speed i just do whatever i got minus the beginning speed?
anonymous
  • anonymous
Indeed.
anonymous
  • anonymous
for the power applied which f *x do i use? the new one?

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