Principles of Physics: A Calculus-Based Text
Principles of Physics: A Calculus-Based Text
5th Edition
ISBN: 9781133104261
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 7, Problem 24P

(a)

To determine

To determine: The speed of the sky diver when he lands on the ground.

(a)

Expert Solution
Check Mark

Answer to Problem 24P

Answer: The speed of the sky diver when he lands on the ground is 24.9m/s .

Explanation of Solution

Explanation:

Given information:

An 80.0kg sky diver jumps out of a balloon at an altitude of 1000m and opens his parachute at an altitude of 200m . Total retarding force on the skydiver with the parachute closed is 50.0N and with the parachute open is 3600N .

Formula to calculate the speed of the sky diver by conservation of energy is,

K1+U1+ΔE=K2+U2 (I)

K1 is the kinetic energy of the sky diver before jump.

K2 is the kinetic energy of the sky diver when he lands at ground.

U1 is the potential energy of the sky diver before jump.

U2 is the potential energy the sky diver to reached the ground.

ΔE is the total mechanical energy.

The initial kinetic energy of the sky diver is zero because sky diver is initially at rest and final potential energy of the sky diver is zero because the distance is zero.

Formula to calculate the potential energy of the sky diver before jump is,

U1=mgh

h is the distance between ground and sky diver jump.

m is the mass of the sky diver.

g is the acceleration due to gravity.

Formula to calculate the total mechanical energy of the system is,

ΔE=F1Δx1+F2Δx2

F1 is the force acting on the sky diver Δx1 when the parachute is closed.

F2 is the force acting on the sky diver when the parachute is open.

Δx1 is the distance covered by the sky diver when the parachute is closed.

Δx2 is the distance covered by the sky diver when the parachute is open.

Formula to calculate the kinetic energy of the sky diver when he lands on ground is,

K2=12mv2

v is the velocity of the sky diver when he lands a ground.

Substitute 12mv2 for K2 , F1Δx1+F2Δx2 for ΔE , mgh for U1 , 0 for U2 and 0 for K1 in equation (I).

(0+mgh(F1Δx1+F2Δx2))=(12mv2+0) (II)

The force F1 and F2 is negative because direction is opposite to the weight of the sky diver.

Substitute 80.0kg for m , 1000m for h , 800m for Δx1 , 200m for Δx2 , 50N for F1 , 3600N for F2 and 9.81m/s2 for g to find v .

(0+80.0kg×9.81m/s2×1000m(50N×800m+3600N×200m))=(1280.0kg×v2+0)v=24.9m/s

Conclusion:

Therefore, the speed of the sky diver when he lands on the ground is 24.9m/s .

(b)

To determine

To explain: Whether the sky diver will be injured or not.

(b)

Expert Solution
Check Mark

Answer to Problem 24P

Answer: Therefore, the sky diver will be injured because his speed is very high.

Explanation of Solution

Explanation:

Given information:

An 80.0kg sky diver jumps out of a balloon at an altitude of 1000m and opens his parachute at an altitude of 200m . Total retarding force on the skydiver with the parachute closed is 50.0N and with the parachute open is 3600N .

Yes, the sky diver will be injured because speed of the sky diver is covered the distance approx 25m in one second. This speed is very high as compared to normal speed so the force of impact is so high when he lands on the ground that he will be injured.

Conclusion:

Therefore, the sky diver will be injured because his speed is very high.

(c)

To determine

To determine: The height at which parachute should be open if the final speed of the sky diver when he hits the ground is 5.00m/s .

(c)

Expert Solution
Check Mark

Answer to Problem 24P

Answer: The height parachute should be open is 219m if the final speed of the sky diver when he hits the ground is 5.00m/s .

Explanation of Solution

Explanation:

Given information:

An 80.0kg sky diver jumps out of a balloon at an altitude of 1000m and opens his parachute at an altitude of 200m .

From equation (II),

(0+mgh(F1Δx1+F2Δx2))=(12mv2+0)

Assume Δx2=x so Δx1=1000mx .

Substitute 80.0kg for m , 1000m for h , 1000mx for Δx1 , x for Δx2 , 50N for F1 , 3600N for F2 , 5.0m/s for v and 9.81m/s2 for g to find x .

(0+80.0kg×9.81m/s2×1000m(50N×(1000mx)3600N×x))=(12×80.0kg×(5.0m/s)2+0)784800N-m50000N-m+(50N)x(3600N)x=1000N-m(3350N)x=733800N-mx=219m

Conclusion:

Therefore, height parachute should be open is 219m if the final speed of the sky diver when he hits the ground is 5.00m/s .

(d)

To determine

To determine: The assumption that the total retarding force is constant.

(d)

Expert Solution
Check Mark

Answer to Problem 24P

Answer: Therefore, as the density of air changes with the altitude therefore, the assumption that the retarding force is constant is not realistic.

Explanation of Solution

Explanation:

Given information:

An 80.0kg sky diver jumps out of a balloon at an altitude of 1000m and opens his parachute at an altitude of 200m .

The assumption that total retarding force is constant is not realistic because the air density changes with the change in altitude. Retarding force is proportional to the density of the air so, with the change in the density of air, the retarding force also changes with the altitude.

Conclusion:

Therefore, as the density of air changes with the altitude therefore, the assumption that the retarding force is constant is not realistic.

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Chapter 7 Solutions

Principles of Physics: A Calculus-Based Text

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