## anonymous 3 years ago The speed of a cycle increases from 2.5 m/s to 5 m/s. The total mass of the cyclist including the cycle is 120 kg. Calculate the work done by the cyclist to increase the speed.

1. shamim

i m trying to solve this problem

2. shamim

here u=2.5 v=5 m=120kg work done W=?

3. shamim

$W=FS=maS$

4. shamim

again we know$v ^{2}=u ^{2}+2aS$

5. shamim

$or, 2aS=v ^{2}-u ^{2}$

6. shamim

$or, aS=\frac{ v ^{2}-u ^{2} }{ 2 }$

7. shamim

putting this value of aS in the 1st equation we get

8. shamim

$W=m(\frac{ v ^{2}-u ^{2} }{ 2 })$

9. shamim

$or, W=120 \times (\frac{ 5^{2}-2.5^{2} }{ 2 })$

10. shamim

or, W=1125 joule

11. shamim

r u following me

12. anonymous

Use work-energy theorem, and write: $$W=m(\Large\frac{ v ^{2}-u ^{2} }{ 2 })$$ directly. This is more general than the derivation provided by shamim, as it will be true even for non-uniform acceleration. But this will only be true if the cyclist is on a horizontal road.