Answer:
a) -6,750 N b) -1.35*10⁶ N c) 11.5 times d) 2,295 times
Explanation:
a) In order to answer the question, we can apply Newton's 2nd Law to the person, as follows:
Fnet = m*a
Assuming that the deceleration is uniform, we can use any of the kinematic equations.
In this particular case, examining the givens that we have (the final speed (which is 0), the initial speed (impact speed), and the distance of deceleration (1 m) the most useful equation is the following:
vf²-v₀² = 2*a*d
Replacing by the givens, and solving for a, we get:
a = (15m/s)² / 2* (1 m) = -112.5 m/s²
With this value of a, we can get the net force F:
F = 60 kg* (-112.5 m/s²) = -6,750 N
b) For this part, same reasoning applies, the only difference being the deceleration distance, which for this case is only 5 mm.
We can apply the same kinematic equation:
vf²-v₀² = 2*a*d
Once again, replacing by the givens, and solving for a, we get:
a = (15m/s)² / 2* (0.005 m) = -22,500 m/s²
With this value of a, we can get the net force F:
F = 60 kg* (-22,500 m/s²) = -1.35*10⁶ N
c) If we compare the forces that we got above with the weight of the person (which is the same to compare the acceleration with g), we have, for the first case (person restrained) a value of approximately 12 g, but for the unrestrained case, we got a value of 2,295 g!
