How much work should be done on a bicycle of mass 20 kg to increase its speed from `2 ms^(-1)` to `5ms^(-1)` ? (Ignore air resistance and friction).

Explanation

We know that whenever work is done, an equal amount of energy is used up. So, the work done in this case will be equal to the change in kinetic energy of bicycle when its speed changes from `2 ms^(-1)` to `5 ms^(-1)`.
(a) In the first case:
Mass of bicycle, m = 20 kg
And, Speed of bicycle, `v = 2 ms^(-1)`
So, Kinetic energy, `E_(k) = (1)/(2)mv^(2)`
`= (1)/(2) xx 20 xx (2)^(2)`
`= 10 xx 4`
= 40 J
(b) In the second case :
Mass of bicycle, m = 20 kg
And, Speed of bicycle, ` v = 5 ms^(-1)`
So, Kinetic energy, `E_(k) = (1)/(2)mv^(2)`
`= (1)/(2) xx 20 xx (5)^(2)`
`= 10 xx 25`
= 250 J
Now, Work done = Change in kinetic energy
= 250 - 40
= 210 J
Thus, the work done is 210 joules.