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Chapter 4: Problem 42

A parachutist relies on air resistance (mainly on her parachute) to decrease her downward velocity. She and her parachute have a mass of \(55.0 \mathrm{~kg},\) and at a particular moment air resistance exerts a total upward force of \(620 \mathrm{~N}\) on her and her parachute. (a) What is the weight of the parachutist? (b) Draw a free-body diagram for the parachutist (see Section 4.6). Use that diagram to calculate the net force on the parachutist. Is the net force upward or downward? (c) What is the acceleration (magnitude and direction) of the parachutist?

Short Answer

Expert verified
(a) 539 N, (b) 81 N upward, (c) 1.47 m/s虏 upward.

Step by step solution

01

Calculate the Weight of the Parachutist

The weight of the parachutist can be calculated using the formula for weight, which is the force due to gravity: \[ W = mg \] where \( W \) is the weight, \( m \) is the mass of the parachutist, and \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \)).Substituting in the given values,\[ W = 55.0 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 539 \, \text{N} \]Therefore, the weight of the parachutist is \( 539 \, \text{N} \).
02

Analyze the Free-Body Diagram

In a free-body diagram, the forces acting on the parachutist include the downward gravitational force (weight) of \( 539 \, \text{N} \) and the upward air resistance force of \( 620 \, \text{N} \). The net force \( F_{\text{net}} \) is the sum of these forces, which can be calculated as:\[ F_{\text{net}} = F_{\text{up}} - W \]Substitute the known values:\[ F_{\text{net}} = 620 \, \text{N} - 539 \, \text{N} = 81 \, \text{N} \]Thus, the net force on the parachutist is \( 81 \, \text{N} \) in the upward direction.
03

Calculate the Acceleration of the Parachutist

The acceleration can be found using Newton's second law of motion:\[ F_{\text{net}} = ma \]Solving for \( a \), we get:\[ a = \frac{F_{\text{net}}}{m} \]Substitute the given values:\[ a = \frac{81 \, \text{N}}{55.0 \, \text{kg}} \approx 1.47 \, \text{m/s}^2 \]Therefore, the magnitude of the acceleration is approximately \( 1.47 \, \text{m/s}^2 \) and its direction is upward.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Free-body diagram
A free-body diagram is a simple representation of an object and all the forces acting on it.
Imagine the parachutist as a dot in the middle of your paper. Every force acting on her, we draw as an arrow.
The direction of the arrow shows where the force is heading.
  • A downward arrow represents the gravitational force or her weight.
  • An upward arrow shows the air resistance acting on her and her parachute.
For this exercise, her weight is depicted as a larger downward arrow with a magnitude of 539 N, and the upward air resistance is an even larger arrow of 620 N.
By drawing these forces, we get a clear picture of how they play against each other to affect her movement.
The net effect of these forces determines how the parachutist moves.
Newton's second law
Newton's second law of motion is a cornerstone of physics, especially when dealing with moving objects.
It states that the acceleration of an object depends on the net force acting on it and its mass.
The formula is straightforward:
\[ F_{\text{net}} = ma \]Where:
  • \( F_{\text{net}} \) is the net force acting on the object.
  • \( m \) is the mass.
  • \( a \) is the acceleration of the object.
This law helps us understand that a greater force results in a greater acceleration, yet taking into account the object's mass.
For the parachutist, this concept allows us to calculate her upward acceleration as influenced by both air resistance and gravity.
Air resistance
Air resistance, also known as drag, is an important force acting on objects moving through air, like a parachutist.
It opposes gravity by acting upwards and can change the speed at which an object falls.
The parachute significantly increases air resistance, which helps slow the parachutist鈥檚 descent.
  • In our problem, the air resistance is 620 N acting upward.
This force plays a vital role in safely reducing the parachutist's speed, allowing for a more controlled and gentle landing.
Calculating how this force competes with weight gives us the net force that influences overall movement.
Weight calculation
Weight is the force exerted on an object due to gravity and can be calculated using a simple equation.
For any object near Earth鈥檚 surface, its weight (\( W \)) can be found with:
\[ W = mg \]Here:
  • \( m \) is the mass of the object, \( 55.0 \, \text{kg} \) for the parachutist.
  • \( g \) is acceleration due to gravity, approximately \( 9.8 \, \text{m/s}^2 \).
  • Thus, her weight calculates to \( 539 \, \text{N} \).
This calculation is foundational for understanding how gravity contributes to the forces acting on the parachutist compared to air resistance.
Net force
Net force is the overall force acting on an object once all individual forces are summed up.
To find the net force in the parachutist scenario, we consider both opposing forces: weight and air resistance.
The equation used is:
\[ F_{\text{net}} = F_{\text{up}} - W \]Substituting values:
  • \( F_{\text{up}} = 620 \, \text{N} \)
  • \( W = 539 \, \text{N} \)
  • The net force results in \( 81 \, \text{N} \) upward.
Finding the net force gives insight into the direction and magnitude of the parachutist's movement.
Knowing it鈥檚 upward means she鈥檚 decelerating at a controlled rate, not plummeting to the ground.
Acceleration calculation
The acceleration of an object tells us how its velocity changes over time.
From Newton's second law, we derive:
\[ a = \frac{F_{\text{net}}}{m} \]Using the parachutist's values:
  • Net force \( (F_{\text{net}}) = 81 \, \text{N} \)
  • Mass \( (m) = 55.0 \, \text{kg} \)
  • Acceleration \( (a) = \frac{81 \, \text{N}}{55.0 \, \text{kg}} \approx 1.47 \, \text{m/s}^2 \)
The result shows a steady upward acceleration.
This decrease in velocity ensures the parachutist enjoys a safe descent instead of a rapid fall.

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Most popular questions from this chapter

Human biomechanics. The fastest pitched baseball was measured at \(46 \mathrm{~m} / \mathrm{s}\). Typically, a baseball has a mass of \(145 \mathrm{~g}\). If the pitcher exerted his force (assumed to be horizontal and constant) over a distance of \(1.0 \mathrm{~m},\) (a) what force did he produce on the ball during this record-setting pitch? (b) Make free-body diagrams of the ball during the pitch and just after it has left the pitcher's hand.

A chair of mass \(12.0 \mathrm{~kg}\) is sitting on the horizontal floor; the floor is not frictionless. You push on the chair with a force \(F=40.0 \mathrm{~N}\) that is directed at an angle of \(37.0^{\circ}\) below the horizontal, and the chair slides along the floor. (a) Draw a clearly labeled free-body diagram for the chair. (b) Use your diagram and Newton's laws to calculate the normal force that the floor exerts on the chair.

Jumping to the ground. \(\mathrm{A} 75.0 \mathrm{~kg}\) man steps off a platform \(3.10 \mathrm{~m}\) above the ground. He keeps his legs straight as he falls, but at the moment his feet touch the ground his knees begin to bend, and, treated as a particle, he moves an additional \(0.60 \mathrm{~m}\) before coming to rest. (a) What is his speed at the instant his feet touch the ground? (b) Treating him as a particle, what are the magnitude and direction of his acceleration as he slows down if the acceleration is constant? (c) Draw a free-body diagram of this man as he is slowing down. (d) Use Newton's laws and the results of part (b) to calculate the force the ground exerts on him while he is slowing down. Express this force in newtons and also as a multiple of the man's weight. (e) What are the magnitude and direction of the reaction force to the force you found in part (c)?

Planet X! When venturing forth on Planet X, you throw a \(5.24 \mathrm{~kg}\) rock upward at \(13.0 \mathrm{~m} / \mathrm{s}\) and find that it returns to the same level 1.51 s later. What does the rock weigh on Planet \(X ?\)

You walk into an elevator, step onto a scale, and push the "down" button to go directly from the tenth floor to the first floor. You also recall that your weight is \(625 \mathrm{~N}\). Start each of the following parts with a free- body diagram. (a) If the elevator has an initial acceleration of magnitude \(2.50 \mathrm{~m} / \mathrm{s}^{2},\) what does the scale read? (b) What does the scale read after the elevator reaches it final speed as it heads to the bottom floor?
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