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Refer to the diagram shown below.
W = 87.5 N, the weight of the sandwiched board.
μ = 0.622, the static coefficient of friction.
From the free body diagram of the sandwiched board, obtain
2μF = W
F = W/(2μ) = 87.5/(2*0.622) = 70.34 N
Answer: 70.34 N
W = 87.5 N, the weight of the sandwiched board.
μ = 0.622, the static coefficient of friction.
From the free body diagram of the sandwiched board, obtain
2μF = W
F = W/(2μ) = 87.5/(2*0.622) = 70.34 N
Answer: 70.34 N
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 +
u = initial velocity = 0
a = constant acceleration
60 =
60.84 a = 60 * 2
a = 1.97
For a = 1.97 and t = 2.5 s
d =
Here also u is '0' because the distance is being measured from the starting point.
d =
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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