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 = `6x(5//18)`
` = 5//3` m/s
Final velocity, `v = 12km//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) mv^(2)`
`= (1)/(2) (90) (10//3)^(2)`
` = (1)/(2) (90) (10//3) (10// 3)`
` = 500J`
THe work done by the cyclist = Change in kinetic energy = `K.E_((f)) - K.E_((i))`
` = 500J -125 J = 375J`