Answer :
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 + [tex]\frac{1}{2}at^{2}[/tex]
u = initial velocity = 0
a = constant acceleration
60 = [tex]\frac{1}{2}a7.8^{2}[/tex]
60.84 a = 60 * 2
a = 1.97 [tex]m/s^{2}[/tex]
For a = 1.97 [tex]m/s^{2}[/tex] and t = 2.5 s
d = [tex]\frac{1}{2}at^{2}[/tex]
Here also u is '0' because the distance is being measured from the starting point.
d = [tex]\frac{1}{2} *1.97*(2.5)^{2}[/tex]
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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