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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.
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.

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shamim
 3 years ago
Best ResponseYou've already chosen the best response.1i m trying to solve this problem

shamim
 3 years ago
Best ResponseYou've already chosen the best response.1here u=2.5 v=5 m=120kg work done W=?

shamim
 3 years ago
Best ResponseYou've already chosen the best response.1again we know\[v ^{2}=u ^{2}+2aS\]

shamim
 3 years ago
Best ResponseYou've already chosen the best response.1\[or, 2aS=v ^{2}u ^{2}\]

shamim
 3 years ago
Best ResponseYou've already chosen the best response.1\[or, aS=\frac{ v ^{2}u ^{2} }{ 2 }\]

shamim
 3 years ago
Best ResponseYou've already chosen the best response.1putting this value of aS in the 1st equation we get

shamim
 3 years ago
Best ResponseYou've already chosen the best response.1\[W=m(\frac{ v ^{2}u ^{2} }{ 2 })\]

shamim
 3 years ago
Best ResponseYou've already chosen the best response.1\[or, W=120 \times (\frac{ 5^{2}2.5^{2} }{ 2 })\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Use workenergy 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 nonuniform acceleration. But this will only be true if the cyclist is on a horizontal road.
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