1) A 58kg sprinter, starting from rest, runs 55m in 7.8s at constant acceleratio

Question

1) A 58kg sprinter, starting from rest, runs 55m in 7.8s at constant acceleration.

A) How much work does the sprinter do during the first 2.5s of his run?

B) How much work does the sprinter do during the final 2.5s of his run?

2) A 28.0kg child is on a swing that hangs from 3.00-m-long chains, as shown in (Figure 1) .

B) From her maximum angle of 45 degrees, the child swings back to the bottom of the arc. While swinging back, a gust of wind hits the child and she returns to the bottom with a speed that is 1.6 times larger than your answer to Part A. How much work was done by the wind during this motion?

Explanation / Answer

x = u*t + 0.5a*t^2
V0 = 0 ---Initial velocity.
x = 55 meter
t = 7.8 seconds
Therefore,
a = 55/(0.5)(7.8)^2
a = 1.80 m/s^2

F = m*a
m = 58 kg
F = 58(1.80)
F = 104.4 Newton

a). initial velocity u= 0
At 0 - 2.5 seconds,

x = U*t+ 0.5at^2 (as u=0)
x = 0.5at^2
x = 0.5(1.80(2.5)^2)
x = 5.625 meters

a) . work W =F*displacement =  104.4*5.625 = 587.25 J

b) for final 2.5 second means t = 5.3 to 7.8 seconds.

X1 = 0.5(1.80(5.3)^2) = 25.81m

X2 =  0.5(1.80(7.8)^2) = 54.756 m

so net displacement in final 2.5 sec =  54.756-25.81 = 28.946m

so work in final 2,5 sec

W =F*d = 104.4*28.946 = 3021.96N

second part give figure..

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