The answer for the following problem is mentioned below.
- Therefore the work done to increase the velocity is 3640 J.
Explanation:
Given:
initial speed (u) =5 m/s
final speed (v) = 9 m/s
mass of the cyclist (m) =130 kg
To solve:
Wok needed to increase the velocity (W)
We know;
initial kinetic energy ([tex]KE_{1}[/tex]) = [tex]\frac{1}{2}[/tex] × m × (u²)
initial kinetic energy ([tex]KE_{1}[/tex]) = [tex]\frac{1}{2}[/tex] × 130 × ( 5 ×5 )
initial kinetic energy ([tex]KE_{1}[/tex]) = 65 × 25 =1625 J
final kinetic energy([tex]KE_{2}[/tex]) = [tex]\frac{1}{2}[/tex] × m × (v²)
final kinetic energy([tex]KE_{2}[/tex]) = [tex]\frac{1}{2}[/tex] × 130 × ( 9×9 )
final kinetic energy([tex]KE_{2}[/tex]) = 81 × 65
final kinetic energy([tex]KE_{2}[/tex]) = 5265 J
Work done (W) =ΔK.E =([tex]KE_{2}[/tex]) - ([tex]KE_{1}[/tex])
Work done (W) = 5265 - 1625 =3640 J
Work done (W) = 3640 J
Therefore the work done to increase the velocity is 3640 J.
