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 = 6 km/h = `6 xx (5//18)`
= 5/3 m/s
Final velocity , `v = 12 km//h = 12 xx (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_((i))=1/2 mv^(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 = 375 J ` .