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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.
 one year ago
 one year 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.
 one year ago
 one year ago

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shamimBest ResponseYou've already chosen the best response.1
i m trying to solve this problem
 one year ago

shamimBest ResponseYou've already chosen the best response.1
here u=2.5 v=5 m=120kg work done W=?
 one year ago

shamimBest ResponseYou've already chosen the best response.1
again we know\[v ^{2}=u ^{2}+2aS\]
 one year ago

shamimBest ResponseYou've already chosen the best response.1
\[or, 2aS=v ^{2}u ^{2}\]
 one year ago

shamimBest ResponseYou've already chosen the best response.1
\[or, aS=\frac{ v ^{2}u ^{2} }{ 2 }\]
 one year ago

shamimBest ResponseYou've already chosen the best response.1
putting this value of aS in the 1st equation we get
 one year ago

shamimBest ResponseYou've already chosen the best response.1
\[W=m(\frac{ v ^{2}u ^{2} }{ 2 })\]
 one year ago

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

VincentLyon.FrBest ResponseYou've already chosen the best response.0
Use 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.
 one year ago
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