physics chapter 5 homework

Check Your Understanding

14 N, 56 ° 56 ° measured from the positive x -axis

a. Their weight acts downward, and the force of air resistance with the parachute acts upward. b. neither; the forces are equal in magnitude

0.1 m/s 2 0.1 m/s 2

40 m/s 2 40 m/s 2

a. 159.0 i ^ + 770.0 j ^ N 159.0 i ^ + 770.0 j ^ N ; b. 0.1590 i ^ + 0.7700 j ^ N 0.1590 i ^ + 0.7700 j ^ N

a = 2.78 m/s 2 a = 2.78 m/s 2

a. 3.0 m / s 2 3.0 m / s 2 ; b. 18 N

a. 1.7 m/s 2 ; 1.7 m/s 2 ; b. 1.3 m/s 2 1.3 m/s 2

6.0 × 10 2 6.0 × 10 2 N

Conceptual Questions

Forces are directional and have magnitude.

The cupcake velocity before the braking action was the same as that of the car. Therefore, the cupcakes were unrestricted bodies in motion, and when the car suddenly stopped, the cupcakes kept moving forward according to Newton’s first law.

No. If the force were zero at this point, then there would be nothing to change the object’s momentary zero velocity. Since we do not observe the object hanging motionless in the air, the force could not be zero.

The astronaut is truly weightless in the location described, because there is no large body (planet or star) nearby to exert a gravitational force. Her mass is 70 kg regardless of where she is located.

The force you exert (a contact force equal in magnitude to your weight) is small. Earth is extremely massive by comparison. Thus, the acceleration of Earth would be incredibly small. To see this, use Newton’s second law to calculate the acceleration you would cause if your weight is 600.0 N and the mass of Earth is 6.00 × 10 24 kg 6.00 × 10 24 kg .

a. action: Earth pulls on the Moon, reaction: Moon pulls on Earth; b. action: foot applies force to ball, reaction: ball applies force to foot; c. action: rocket pushes on gas, reaction: gas pushes back on rocket; d. action: car tires push backward on road, reaction: road pushes forward on tires; e. action: jumper pushes down on ground, reaction: ground pushes up on jumper; f. action: gun pushes forward on bullet, reaction: bullet pushes backward on gun.

a. The rifle (the shell supported by the rifle) exerts a force to expel the bullet; the reaction to this force is the force that the bullet exerts on the rifle (shell) in opposite direction. b. In a recoilless rifle, the shell is not secured in the rifle; hence, as the bullet is pushed to move forward, the shell is pushed to eject from the opposite end of the barrel. c. It is not safe to stand behind a recoilless rifle.

a. Yes, the force can be acting to the left; the particle would experience acceleration opposite to the motion and lose speed. B. Yes, the force can be acting downward because its weight acts downward even as it moves to the right.

two forces of different types: weight acting downward and normal force acting upward

a. F → net = 5.0 i ^ + 10.0 j ^ N F → net = 5.0 i ^ + 10.0 j ^ N ; b. the magnitude is F net = 11 N F net = 11 N , and the direction is θ = 63 ° θ = 63 °

a. F → net = 660.0 i ^ + 150.0 j ^ N F → net = 660.0 i ^ + 150.0 j ^ N ; b. F net = 676.6 N F net = 676.6 N at θ = 12.8 ° θ = 12.8 ° from David’s rope

a. F → net = 95.0 i ^ + 283 j ^ N F → net = 95.0 i ^ + 283 j ^ N ; b. 299 N at 71 ° 71 ° north of east; c. F → DS = − ( 95.0 i ^ + 283 j ^ ) N F → DS = − ( 95.0 i ^ + 283 j ^ ) N

Running from rest, the sprinter attains a velocity of v = 12.96 m/s v = 12.96 m/s , at end of acceleration. We find the time for acceleration using x = 20.00 m = 0 + 0.5 a t 1 2 x = 20.00 m = 0 + 0.5 a t 1 2 , or t 1 = 3.086 s. t 1 = 3.086 s. For maintained velocity, x 2 = v t 2 x 2 = v t 2 , or t 2 = x 2 / v = 80.00 m / 12.96 m / s = 6.173 s t 2 = x 2 / v = 80.00 m / 12.96 m / s = 6.173 s . Total time = 9.259 s Total time = 9.259 s .

a. m = 56.0 kg m = 56.0 kg ; b. a meas = a astro + a ship , where a ship = m astro a astro m ship a meas = a astro + a ship , where a ship = m astro a astro m ship ; c. If the force could be exerted on the astronaut by another source (other than the spaceship), then the spaceship would not experience a recoil.

F net = 4.12 × 10 5 N F net = 4.12 × 10 5 N

a = 253 m/s 2 a = 253 m/s 2

F net = F − f = m a ⇒ F = 1.26 × 10 3 N F net = F − f = m a ⇒ F = 1.26 × 10 3 N

v 2 = v 0 2 + 2 a x ⇒ a = − 7.80 m/s 2 F net = −7.80 × 10 3 N v 2 = v 0 2 + 2 a x ⇒ a = − 7.80 m/s 2 F net = −7.80 × 10 3 N

a. F → net = m a → ⇒ a → = 9.0 i ^ m/s 2 F → net = m a → ⇒ a → = 9.0 i ^ m/s 2 ; b. The acceleration has magnitude 9.0 m/s 2 9.0 m/s 2 , so x = 110 m x = 110 m .

1.6 i ^ − 0.8 j ^ m/s 2 1.6 i ^ − 0.8 j ^ m/s 2

a. w Moon = m g Moon m = 150 kg w Earth = 1.5 × 10 3 N w Moon = m g Moon m = 150 kg w Earth = 1.5 × 10 3 N ; b. Mass does not change, so the suited astronaut’s mass on both Earth and the Moon is 150 kg. 150 kg.

a. F h = 3.68 × 10 3 N and w = 7.35 × 10 2 N F h w = 5.00 times greater than weight F h = 3.68 × 10 3 N and w = 7.35 × 10 2 N F h w = 5.00 times greater than weight ; b. F net = 3750 N θ = 11.3 ° from horizontal F net = 3750 N θ = 11.3 ° from horizontal

w = 19.6 N F net = 5.40 N F net = m a ⇒ a = 2.70 m/s 2 w = 19.6 N F net = 5.40 N F net = m a ⇒ a = 2.70 m/s 2

a. F net = 2.64 × 10 7 N; F net = 2.64 × 10 7 N; b. The force exerted on the ship is also 2.64 × 10 7 N 2.64 × 10 7 N because it is opposite the shell’s direction of motion.

Because the weight of the history book is the force exerted by Earth on the history book, we represent it as F → EH = −14 j ^ N . F → EH = −14 j ^ N . Aside from this, the history book interacts only with the physics book. Because the acceleration of the history book is zero, the net force on it is zero by Newton’s second law: F → PH + F → EH = 0 → , F → PH + F → EH = 0 → , where F → PH F → PH is the force exerted by the physics book on the history book. Thus, F → PH = − F → EH = − ( −14 j ^ ) N = 14 j ^ N . F → PH = − F → EH = − ( −14 j ^ ) N = 14 j ^ N . We find that the physics book exerts an upward force of magnitude 14 N on the history book. The physics book has three forces exerted on it: F → EP F → EP due to Earth, F → HP F → HP due to the history book, and F → DP F → DP due to the desktop. Since the physics book weighs 18 N, F → EP = −18 j ^ N . F → EP = −18 j ^ N . From Newton’s third law, F → HP = − F → PH , F → HP = − F → PH , so F → HP = −14 j ^ N . F → HP = −14 j ^ N . Newton’s second law applied to the physics book gives ∑ F → = 0 → , ∑ F → = 0 → , or F → DP + F → EP + F → HP = 0 → , F → DP + F → EP + F → HP = 0 → , so F → DP = − ( −18 j ^ ) − ( −14 j ^ ) = 32 j ^ N . F → DP = − ( −18 j ^ ) − ( −14 j ^ ) = 32 j ^ N . The desk exerts an upward force of 32 N on the physics book. To arrive at this solution, we apply Newton’s second law twice and Newton’s third law once.

a. The free-body diagram of pulley 4:

b. T = m g , F = 2 T cos θ = 2 m g cos θ T = m g , F = 2 T cos θ = 2 m g cos θ

a. 65 N b. 1.22 × 10 4 m/s 2 1.22 × 10 4 m/s 2

a. T = 1.96 × 10 −4 N; T = 1.96 × 10 −4 N; b. T ′ = 4.71 × 10 −4 N T ′ T = 2.40 times the tension in the vertical strand T ′ = 4.71 × 10 −4 N T ′ T = 2.40 times the tension in the vertical strand

F y net = F ⊥ − 2 T sin θ = 0 F ⊥ = 2 T sin θ T = F ⊥ 2 sin θ F y net = F ⊥ − 2 T sin θ = 0 F ⊥ = 2 T sin θ T = F ⊥ 2 sin θ

a. see Example 5.13 ; b. 1.5 N; c. 15 N

a. 5.6 kg; b. 55 N; c. T 2 = 60 N T 2 = 60 N ; d.

a. 4.9 m/s 2 4.9 m/s 2 , 17 N; b. 9.8 N

Additional Problems

a. F net = m ( v 2 − v 0 2 ) 2 x F net = m ( v 2 − v 0 2 ) 2 x ; b. 2590 N

F → net = F → 1 + F → 2 + F → 3 = ( 6.02 i ^ + 14.0 j ^ ) N F → net = m a → ⇒ a → = F → net m = 6.02 i ^ + 14.0 j ^ N 10.0 kg = ( 0.602 i ^ + 1.40 j ^ ) m/s 2 F → net = F → 1 + F → 2 + F → 3 = ( 6.02 i ^ + 14.0 j ^ ) N F → net = m a → ⇒ a → = F → net m = 6.02 i ^ + 14.0 j ^ N 10.0 kg = ( 0.602 i ^ + 1.40 j ^ ) m/s 2

F → net = F → A + F → B F → net = A i ^ + ( −1.41 A i ^ − 1.41 A j ^ ) F → net = A ( −0.41 i ^ − 1.41 j ^ ) θ = 254 ° F → net = F → A + F → B F → net = A i ^ + ( −1.41 A i ^ − 1.41 A j ^ ) F → net = A ( −0.41 i ^ − 1.41 j ^ ) θ = 254 ° (We add 180 ° 180 ° , because the angle is in quadrant IV.)

F = 2 m k 2 x 2 F = 2 m k 2 x 2 ; First, take the derivative of the velocity function to obtain a = 2 k x v = 2 k x ( k x 2 ) = 2 k 2 x 3 a = 2 k x v = 2 k x ( k x 2 ) = 2 k 2 x 3 . Then apply Newton’s second law F = m a = 2 m k 2 x 2 F = m a = 2 m k 2 x 2 .

a. For box A, N A = m g N A = m g and N B = m g cos θ N B = m g cos θ ; b. N A > N B N A > N B because for θ < 90 ° θ < 90 ° , cos θ < 1 cos θ < 1 ; c. N A > N B N A > N B when θ = 10 ° θ = 10 °

a. 8.66 N; b. 0.433 m

0.40 or 40%

Challenge Problems

; b. No; F → R F → R is not shown, because it would replace F → 1 F → 1 and F → 2 F → 2 . (If we want to show it, we could draw it and then place squiggly lines on F → 1 F → 1 and F → 2 F → 2 to show that they are no longer considered.

a. 14.1 m/s; b. 601 N

F m t 2 F m t 2

a → = −248 i ^ − 433 j ^ m / s 2 a → = −248 i ^ − 433 j ^ m / s 2

0.548 m/s 2 0.548 m/s 2

a. T 1 = 2 m g sin θ T 1 = 2 m g sin θ , T 2 = m g sin ( arctan ( 1 2 tan θ ) ) T 2 = m g sin ( arctan ( 1 2 tan θ ) ) , T 3 = 2 m g tan θ ; T 3 = 2 m g tan θ ; b. ϕ = arctan ( 1 2 tan θ ) ϕ = arctan ( 1 2 tan θ ) ; c. 2.56 ° 2.56 ° ; (d) x = d ( 2 cos θ + 2 cos ( arctan ( 1 2 tan θ ) ) + 1 ) x = d ( 2 cos θ + 2 cos ( arctan ( 1 2 tan θ ) ) + 1 )

a. a → = ( 5.00 m i ^ + 3.00 m j ^ ) m / s 2 ; a → = ( 5.00 m i ^ + 3.00 m j ^ ) m / s 2 ; b. 1.38 kg; c. 21.2 m/s; d. v → = ( 18.1 i ^ + 10.9 j ^ ) m / s 2 v → = ( 18.1 i ^ + 10.9 j ^ ) m / s 2

a. 0.900 i ^ + 0.600 j ^ N 0.900 i ^ + 0.600 j ^ N ; b. 1.08 N

As an Amazon Associate we earn from qualifying purchases.

This book may not be used in the training of large language models or otherwise be ingested into large language models or generative AI offerings without OpenStax's permission.

Want to cite, share, or modify this book? This book uses the Creative Commons Attribution License and you must attribute OpenStax.

Access for free at https://openstax.org/books/university-physics-volume-1/pages/1-introduction

• Authors: William Moebs, Samuel J. Ling, Jeff Sanny
• Publisher/website: OpenStax
• Book title: University Physics Volume 1
• Publication date: Sep 19, 2016
• Location: Houston, Texas
• Book URL: https://openstax.org/books/university-physics-volume-1/pages/1-introduction
• Section URL: https://openstax.org/books/university-physics-volume-1/pages/chapter-5

© Jan 19, 2024 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License . The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.

Chapter 5 Homework

Back to ph 220 homework page, back to dr. donovan's courses page, back to dr. donovan's homepage, physics dept. homepage, nmu homepage.

Want to create or adapt books like this? Learn more about how Pressbooks supports open publishing practices.

5 Chapter 5: Introduction to Forces

Textbook Chapter 5: Newton’s Laws of Motion

Section 5.1: Newton’s First Law

Textbook Section 5.2: Newton’s First Law

Textbook Section 5.4: Mass and Weight

A force is defined as a push or a pull on an object. This might be due to a direct interaction between two objects ( contact force ) or due to a long-range force such as gravity. The contact forces we will consider include the normal, friction, and tension forces.

Forces are vector quantities with both magnitude and direction. Forces are added as vectors.

Newton’s First Law: the law of inertia. An object in motion (has a constant or zero velocity) will remain in motion unless acted upon by an outside force.

Newton’s first law tells us that any change in velocity (magnitude, direction, or both) must be due to a force acting on the object. What forces might act on a book pushed across a table to bring it to a stop? These forces can include friction with both the table and the air (we will address frictional forces in Chapter 6). Our own experience tends to disagree with Newton’s first law because we are always in the presence of frictional forces on Earth. But if you threw a book into the emptiness of space, it would continue to move at the velocity you threw it until it ran into something. There’s nothing to slow it down.

Newton’s first law is only valid in inertial reference frames (where the object is not accelerating – moving at constant velocity or not moving at all). We will only consider these cases.

Section 5.2: Newton’s Second Law

Textbook Section 5.3: Newton’s Second Law

Section 5.3: Newton’s Third Law

Textbook Section 5.5: Newton’s Third Law

Newton’s Third Law: if object A exerts a force on object B, object B exerts an equal and opposite force on object A.

This law refers to the interaction between two objects. If one person pushes something, the object pushes back. You feel this every time you interact with something — heavy objects put up a lot of resistance to motion, and lighter objects put up less resistance to motion, but everything pushes back at least a little bit.

A common mistake people make when interpreting the third law is to think that since the forces are equal and in opposite directions, the reactions of the two objects will be equal. This isn’t the case! If their masses are different, the objects will both have different accelerations. The more massive object has a small acceleration, and the less massive object has a much larger acceleration.

Another common mistake is to think that because two forces are of equal value and in opposite directions, they must be third-law pairs. This isn’t necessarily the case. Third law pairs are between two objects only. If there are more than two objects between these forces, it’s not a third law pair.

Section 5.4: Free Body Diagrams (FBDs)

Textbook Section 5.7: Drawing Free Body Diagrams

For any object, you can draw a free body diagram (FBD). The free body diagram represents the object as a point, and you include all the individual forces acting on the object. You do NOT include the total force acting on the object (which is instead the vector sum of all individual forces) and you do not include forces the object is asserting on other objects. Only forces directly acting on the object at hand are included.

Consider a can of soda resting on your desk. What forces are present? Gravity is pulling down on the soda can, and the desk is pushing up on the soda can. The sum of these two forces must be equal to zero if the soda can is not accelerating (and it’s not, it’s sitting still). Thus, these forces are of equal and opposite magnitudes. Are they third law pairs? No! One force is between the soda can and the Earth, and the other force is between the soda can and the desk. So where are our third-law pairs? You will never have a third law pair on a FBD. Since the FBD only includes forces acting on the object, it will not include any forces of the object acting on something else (which must be the case for the third law pairs). The third law pairs to the forces above would be the soda can pulling up on the Earth and the soda can pushing down on the desk. Your resulting FBD for the soda can is given below.

Because the upwards force of the desk on the soda is equal to the downwards force of the Earth on the soda, these two vectors should be given the same magnitude (length) in your drawing. If one force is bigger than another, make sure to indicate that in your FBD by the length of the vectors. Always label your vectors according to which force is acting on the object.

Section 5.5: Normal Forces

Textbook Section 5.6: Common Forces

What do we mean by a normal force? In this case, normal means perpendicular . Whenever you come into contact with an object (pushing on a door to open it, standing on a table or the ground, hitting a baseball with a baseball bat) you exert a normal force on that object (a force perpendicular to the surface of the object).

Think of the surface of any object (door, Earth, table, ball, etc.) like the rubber sheet of a trampoline. When you put an object on a trampoline, the surface deforms. But the trampoline has a restoring force associated with it, which acts like a spring when compressed. When you remove the object from the trampoline surface, the surface goes back to it’s normal state, a flat sheet. A heavier object deforms the surface further, while a light object only deforms the surface a little bit.

This restoring force is our normal force. Any force applied to an object, no matter how hard the surface of the object appears, deforms the surface in some way. This deformation leads to a restoring force that attempts to bring the surface back to where it was. When you push down on the ground, a restoring force called the normal force pushes back up.

There are limits to the normal force. You can only push down so far on some things before the normal force can no longer counter the applied force, and this is when objects break (like a table with an elephant on top).

A normal force always acts perpendicular to the surface which supplies the normal force. This does  not mean the normal force is always perpendicular to the applied force. Consider a box sitting on a ramp. The force of gravity applies straight down, but the normal force from the ramp applies perpendicular to the ramp, not opposite the force of gravity. As a result, the gravitational and normal forces do not cancel out in this example. There is a net force in the direction down the ramp, and the box will experience an acceleration in that direction, which is why it will slide down the ramp.

Section 5.6: Tension Forces

Tension forces refer to forces applied to or by a string, chain, wire, etc. If you are hanging from a rope ladder above the ground, there is a tension force in the rope ladder equal to the force of gravity (assuming you haven’t fallen yet). But put too much force on the ladder (keep adding more weight) and it will eventually break. Everything that can give a tension force has a limit to how much it can take. That’s where the expression “you’re only as strong as your weakest link” comes from — the tension a chain of links can take is limited to the tension that can be applied to the weakest link of the chain. If one breaks, the tension is gone. In all of the problems we do, assume that the tension is equal along the entire rope/chain/etc.

Section 5.7: Solving Force Problems

Draw a picture.

Draw and label all forces acting on the object.

Draw a FBD for your object.

Solve for whatever variable you are looking for using the equations you wrote in the previous step. Keep in mind you might need to use the kinematics equations as well.

In Class Group Problem 5.1:

A person pushes a box across a frictionless surface with a force of 12 N. If the box starts from rest and has reached a velocity of 3.5 m/s after 6.0 s, what is the mass of the box?

Draw a free body diagram (FBD) for the box and label all the forces acting on the box. Is the net force on the box equal to zero? In which direction does the net force point?

In Class Group Problem 5.2:

What is Bill’s mass in kg?

Why don’t we see the Earth moving up to meet us halfway as we fall towards it?

In Class Group Problem 5.3:

How does the force exerted by the mosquito on the car compare to the force exerted by the car on the mosquito?

Consider your answer to the previous question. Think about what happens to the mosquito. Explain why your answer makes sense in terms of Newton’s second law.

What is the change in acceleration of the car as a result?

In Class Group Problem 5.4:

A box is sitting at rest on a frictionless incline as shown below. A rope attached to the box (arrow) keeps it stationary. If the box has a mass of 35 kg, what is the tension in the rope (in other words, what force is it exerting to keep the box from moving)?

In Class Group Problem 5.5:

A 42 kg gymnast is dangling on a massless rope. Draw a FBD for the person and find the tension in the rope.

In Class Group Problem 5.6:

In Class Group Problem 5.7:

In Class Group Problem 5.8:

In Class Group Problem 5.9:

In class group problem 5.10:, in class group problem 5.11:.

While driving in the mountains, you notice that when the freeway goes steeply downhill, there are emergency exits every few miles. These emergency exits are straight dirt ramps which leave the freeway and are sloped uphill. They are designed to stop trucks and cars that lose their breaks on the down hill stretches of the freeway even if the road is covered in ice. You wonder at what angle from the horizontal an emergency exit should rise to stop a 50 ton truck going 70 mph up a ramp 300 ft long, even if the frictional force of the road surface is negligible.

A push or pull on an object.

Objects that come into direct surface-to-surface contact to transmit a force.

Objects do not need to come into direct surface-to-surface contact to transmit these forces. They are instead transmitted by fields, which we'll get into in PHYS 202.

The amount of stuff you have that makes up an object. Mass is measured in kg, and doesn't change when you move an object to another planet or put it in space.

Weight is the amount Earth (or another body) pulls on an object. This is NOT the same thing as mass, since it changes depending on the position of the object (on the Earth vs. the Moon vs. floating in space). Weight is mass x gravity, and is a force.

The gravitational force is the force between two objects that both have mass.

Forces that are equal in magnitude, opposite in direction, and are the opposites of each other - object A acts on object B, so object B acts back on object A.

A diagram that represents an object as a dot, and all the external forces acting on that object as vectors originating from the dot.

A normal force acts perpendicular to the surface causing it.

A force that tries to put something back where it was.

Introductory Physics Resources Copyright © by Adria C Updike is licensed under a Creative Commons Attribution 4.0 International License , except where otherwise noted.

Share This Book