The mass of a cyclist together with the bicycle is 90 kg. Calculate the work done by cyclist if the speed increases from 6km /h to 12 km / h.

Explanation

Mass of cyclist together with bike,
`m = 90 kg`.
Initial velocity, `u = 6km//h = 6xx(5//18)`
`= 5//3 m//s`
Final velocity, `v = 12 km //h = 12xx(5//18)`
`= 10//3 m//s`
Initial kinetic energy
`K.E_((i)) = 1//2 "mu"^2`
`= 1//2 (90) (5//3)^2`
` = 1//2 (90) (5//3) (5/3) = 125 J`
Final kinetic energy,
`K.E (f) = 1//2 m v^2`
`= 1//2 (90) (10//3)^2`
`= 1//2 (90) (10//3)(10//3) = 500 J`
The work done by the cyclist = Change in kinetic energy `= K.E_((f)) - K.E_((i)) = 500 J – 125 J = 375J`