# How much work done by an triathlete who accelerates on the bicycle (combined mass of 105Kg) from 5.0 m/s to 10.0 m/s? The work done by a body is

`W = F d` joules

where `F` is the force acting on the body and `d` is the distance moved

Now, using Newton's second law

`F = ma`

where `m` is the mass of the body in kg and `a` is the acceleration...

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The work done by a body is

`W = F d` joules

where `F` is the force acting on the body and `d` is the distance moved

Now, using Newton's second law

`F = ma`

where `m` is the mass of the body in kg and `a` is the acceleration in m/s^2

Here

`F = 105a` Newtons

Now, the velocity at time `t` is given by

`v = 5 + at`

When the triathlete ` `has accerelated to 10m/s, `v = 10` and

`a = (10-5)/t = 5/t`

We then have that

`F = ma = 105a = 105 (5/t) = 525/t`

The distance travelled by time `t` is given by

`d = 5t + 1/2at^2 = 5t + 1/2(5/t)t^2 = 5t + 2.5t = 7.5t`

Therefore the work done is

`W = Fd = (525/t)(7.5t) = 3937.5` joules

Approved by eNotes Editorial Team