Question

a 58 kg sprinter, starting from rest, runs a 60 m long race in 7.8 s with a constant acceleration. how much work does the sprinter do during the first 2.5 s of the race?

Answer 1

Work done by the sprinter in first 2.5 s will be 702.699 J.

Work done is the product of force and distance covered.

• Mathematically, W = Force * Distance

Mass of the sprinter = 58 kg

Distance covered by the sprinter = s = 60 m

Total time taken by the sprinter = t = 7.8 s

• According to the Equations of motion S = ut + $\frac{1}{2}at^{2}$

u = initial velocity = 0

a = constant acceleration

60 =  $\frac{1}{2}a7.8^{2}$

60.84 a = 60 * 2

a = 1.97 $m/s^{2}$

For a = 1.97 $m/s^{2}$ and t = 2.5 s

d =  $\frac{1}{2}at^{2}$

Here also u is '0' because the distance is being measured from the starting point.

d = $\frac{1}{2} *1.97*(2.5)^{2}$

d = 6.15 m

Work done by the sprinter = F * d

Force = mass * acceleration

F = 58 * 1.97

F = 114.26 N

distance for first 2.5 sec is 6.15 m

Therefore Work done = 6.15 * 114.26

W = 702.699 J

Work done by the sprinter is 702.699 J.

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