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An engine performs 6400 j of work on a motorbike the motorbike and the rider have a combined mass of 200 kg if the bike started at rest what is the speed of the bike after the work is performed

Respuesta :

skyluke89

The work done by the engine is converted into kinetic energy of the motorbike+rider system:

[tex] W=K=\frac{1}{2}mv^2 [/tex]

where

W=6400 J is the work

m=200 kg is the mass of the system

v is the speed acquired by the motorbike


Rearranging the equation and substituting the numbers in, we find

[tex] v=\sqrt{\frac{2W}{m}} =\sqrt{\frac{2(6400 J)}{200 kg}} =8 m/s [/tex]

Eduard22sly

The speed of the bike after the work is performed is 8 m/s

Data obtained from the question

  • Energy (E) = 6400 J
  • Mass (m) = 200 Kg
  • Velocity (v) =?

How to determine the velocity

Energy, mass and velocity of an object is related according to the following equation:

Energy (E) = ½ × mass (m) × Square velocity (v²)

E = ½mv²

With the above formula, we can obtain the velocity of the bike as follow:

E = ½mv²

6400 = ½ × 200 × v²

6400 = 100 × v²

Divide both side by 100

v² = 6400 / 100

v² = 64

Take the square root of both side

v = √64

v = 8 m/s

Learn more about energy:

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