Question

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

Answer 1

Answer:

Explanation:

Answer

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Arnold11

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.

Answer 2

Step by step Explanation:

Given:

Mass(m): $60kg$

Speed(v):$2m/s$

Work Output(W): $1800J$

To find:

Velocity(V)

Solution:

Kinetic energy

                  $KE=\frac{1}{2}mv^2$                     where,m =Mass, v=Speed

substitute the values in the formula,

                $KE=\frac{1}{2}(60)(2)^2$

                        $=\frac{1}{2}(60)(4)$

                        $=\frac{1}{2}(240)$

                $KE=120J$

Total kinetic energy

$KE_{total=KE+W$

           $=120J+1800J$

           $=1920 J$

$KE_{total}=1920 J = \frac{1}{2}\times m\times V^2$

                $1920J=\frac{1}{2}\times 60\times V^2$

                $1920J=30\times V^2$

                  $\frac{1920J}{30} =V^2$

                     $V^2=64$

                      $V=8m/s$

Therefore, the final velocity is   $(8m/s)$.