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sergejj [24]
3 years ago
7

A 60 kg bicyclist going 2 m/s increased his work output by 1,800 J. What was his final velocity?

Physics
2 answers:
Alinara [238K]3 years ago
6 0

<em>8 </em>was the correct answer, thanks

Bas_tet [7]3 years ago
3 0
Bicyclist initial kinetic energy is Ek=(1/2)*m*v² where m is his mass and v is his speed and that is equal to:

Ek=(1/2)*60*2²=120 J.

When we add the increased work output, we get the total kinetic energy:

Ek(total)=Ek+W= 120 J + 1800 J= 1920 J

So Ek(total)=1920 J = (1/2)*m*V² where V is the speed after the bicyclist increased his work output. So lets solve for V:

(1/2)*60*V²=1920

30*V²=1920, we divide by 30,

V²=64, and take the square root of both sides,

V=8 m/s. 

So the speed of the bicyclist after the increased work output is V=8 m/s.
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