hw07 - Problem 9.32(a Calculate the impulse experienced...

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Problem 9.32 (a) Calculate the impulse experienced when a 60 kg person lands on firm ground after jumping from a height of 2.7 m. (b) Estimate the average force exerted on the person’s feet by the ground if the landing is stiff-legged. With stiff legs, assume the body moves 1.9 cm during impact. (b) Estimate the average force exerted on the person’s feet by the ground if the landing is with bent legs. When the legs are bent, assume the body moves about 44cm during impact. Strategy. For Part (a), the Impulse-Momentum Theorem tells us that the impulse experienced by the woman is equal to the change in her momentum. In order to calculate the change in momentum, we will use conservation of energy to find the woman’s momentum just before she hits the ground. For Parts (b) and (c), The Work-Energy Theorem tells us that the net work done on the woman is equal to the change in her kinetic energy, which is zero. In order to calculate the average force exerted on her by the ground, we must evaluate the work done by both gravity and the ground. Solution. There are three “points of interest” in this problem: (1) the instant at which the woman is at a height h above the ground; (2) the instant immediately before the woman hits the ground; and, (3) the instant immediately after the woman has landed and come to rest. We will denote the woman’s energy and momentum at each of the points by E 1 , 2 , 3 and p 1 , 2 , 3 , respectively. Then E 1 = mgh, E 2 = p 2 2 / (2 m ) , E 3 = 0 , p 1 = 0 , p 2 = - p 2 ˆ j , p 3 = 0 , (9.32.1) where m = 60 kg is the mass of the woman and h = 2 . 7 m is the height from which she falls. Here ˆ i , ˆ j , and ˆ k are unit vectors that point in the x -, y -, and z - directions, respectively Because there are no dissipative forces acting on the woman while she falls, energy is conserved between points 1 and 2: E 1 = E 2 p 2 = m p 2 gh. (9.32.2) According to the Impulse-Momentum Theorem, when the woman lands, she experiences an impulse from the ground given by J 23 = p 3 - p 2 J 23 = m p 2 gh ˆ j . (9.32.3) Using g = 9.8 m / s 2 , we find that J 23 = 440 kg · m / s. Here J 23 is the magnitude of J 23 . The work done by gravity is given by W (gravity) 13 = - ( U 3 - U 1 ) = mg ( h + d ) , (9.32.4) where U 1 , 23 denotes the gravitational potential energy at points 1, 2, and 3. Here d is the distance the body moves during the impact. The work done by the ground is W (ground) 13 = Z 3 1 F ( y ) · d x = Z 0 d F ( y ) dy = - Z d 0 F ( y ) dy, (9.32.5) where F ( y ) = F ( y ) ˆ j is the force of the ground on the woman and d x = dx ˆ i + dy ˆ j + dz ˆ k . According to the Work-Energy Theorem, the net work done on the woman is zero. Therefore, W (net) 13 = W (ground) 13 + W (gravity) 13 = K 3 - K 1 = 0 Z d 0 F ( y ) dy = mg ( h + d ) . (9.32.6) From the definition of “average,” we know that the average force is given by h F i = 1 d Z d 0 F ( y ) dy = mg h d + 1 . (9.32.7) The average force is h F i = 8 . 4 × 10 4 N when d = 1.9 cm and h F i = 4200 N when d = 44 cm.

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