physics #4?

[匿名]

2016-11-07 2:59 pm

Question Part
Points
Submissions Used
Biomechanical research has shown that when a 78 kg person is running, the force exerted on each foot as it strikes the ground can be as great as 2300 N.
(a) What is the ratio of the force exerted on the foot by the ground to the person's body weight?

(b) If the only forces acting on the person are (i) the force exerted by the ground and (ii) the person's weight, what is the magnitude and direction of the person's acceleration?
Magnitude m/s2
Direction

(c) If the acceleration found in part (b) acts for 11.0 ms, what is the resulting change in the vertical component of the person's velocity?
m/s

回答 (1)

NCS

2016-11-07 8:17 pm

✔ 最佳答案

(a) 2300N / 78kg*9.8m/s² = 3
or 3:1 expressed as a ratio.

(b) net force F = 2300N - 78kg * 9.8m/s² = 1536 N
and so the acceleration
a = F / m = 1536N / 78kg = 20 m/s², up

(c) Δv = a*t = 20m/s² * 0.011s = 0.22 m/s

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