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# Chapter 7: Basic Physics SE-Force

Difficulty Level: At Grade Created by: CK-12

The Big Idea

Acceleration is caused by force. The more force applied, the greater the acceleration that is produced. Objects with high masses have more inertia and thus resist more strongly changes to its current velocity. In the absence of applied forces, objects simply keep moving at whatever speed they are already going. All forces come in pairs because they arise in the interaction of two objects — you can’t hit without being hit back! In formal language1\begin{align*}^1\end{align*}:

Newton’s 1st\begin{align*}1^{st}\end{align*} Law: Every body continues in its state of rest, or of uniform motion in a right (straight) line, unless it is compelled to change that state by forces impressed upon it.

Newton’s 2nd\begin{align*}2^{nd}\end{align*} Law: The change of motion is proportional to the motive force impressed; and is made in the direction of the right (straight) line in which that force is impressed.

Newton’s 3rd\begin{align*}3^{rd}\end{align*} Law: To every action there is always opposed an equal reaction: or, the mutual actions of two bodies upon each other are always equal, and directed to contrary parts.

1\begin{align*}^1\end{align*}Principia in modern English, Isaac Newton, University of California Press, 1934

Key Concepts

• An object will not change its state of motion (i.e., accelerate) unless an unbalanced force acts on it. Equal and oppositely directed forces do not produce acceleration.
• If no unbalanced force acts on an object the object remains at constant velocity or at rest.
• The force of gravity is called weight and equals mg, where g\begin{align*}g\end{align*} is the acceleration due to gravity of the planet (g=9.8 m/s210 m/s2\begin{align*}g = 9.8 \ m/s^2 \sim 10 \ m/s^2\end{align*}, downward, on Earth).
• Your mass does not change when you move to other planets, because mass is a measure of how much matter your body contains, and not how much gravitational force you feel.
• Newton’s 3rd\begin{align*}3^{rd}\end{align*} Law states for every force there is an equal but opposite reaction force. To distinguish a third law pair from merely oppositely directed pairs is difficult but very important. Third law pairs must obey three rules: they must be of the same type of force, they are exerted on two different objects and they are equal in magnitude and oppositely directed. Example: A block sits on a table. The Earth’s gravity on the block and the force of the table on the block are equal and opposite. But these are not third law pairs, because they are both on the same object and the forces are of different types. The proper third law pairs are: (1) earth’s gravity on block/block’s gravity on earth and (2) table pushes on block/ block pushes on table.
• Pressure is often confused with force. Pressure is force spread out over an area; a small force exerted on a very small area can create a very large pressure; i.e., poke a pin into your arm!
• The force of friction can actually be described in terms of the coefficient of friction, μ\begin{align*}\mu\end{align*}. It is determined experimentally and varies depending upon the two surfaces in contact.
• Static friction acts between two surfaces that are in contact but not in motion with respect to each other. This force prevents objects from sliding. It always opposes potential motion, and it rises in magnitude to a maximum value given by the formula below.2\begin{align*}^2\end{align*}
• Kinetic friction acts between two surfaces that are in contact and in motion with respect to each other. This force reduces the acceleration and it always opposes the direction of motion.2\begin{align*}^2\end{align*}

2\begin{align*}^2\end{align*}Ultimately many of these “contact” forces are due to attractive and repulsive electromagnetic forces between atoms in materials.

Key Equations

• a=Fnetm\begin{align*}a = \frac{F_{net}}{m}\end{align*} ; the acceleration produced depends on the net force on an object and its mass.
or Fnet=Findividual forces=ma\begin{align*}F_{net} = \sum F_{\text{individual forces}} = ma\end{align*} ; the net force is the vector sum of all the forces
Fnet,x=Fxdirection forces=max\begin{align*}F_{net, x} = \sum F_{x - \text{direction forces}} = ma_x\end{align*}
Fnet,y=Fydirection forces=may\begin{align*}F_{net, y} = \sum F_{y- \text{direction forces}} = ma_y\end{align*} ; acting on the object.
• Fg=mg\begin{align*}F_g = mg\end{align*} ; the force of gravity acting on an object, often simply called the “weight” of the object. On Earth, g=9.8 m/s2\begin{align*}g = 9.8 \ m/s^2\end{align*} in the downward direction.
• N\begin{align*}N\end{align*} or FN\begin{align*}F_N\end{align*} ; the normal force is a contact force that acts in a perpendicular direction to a surface.2\begin{align*}^2\end{align*}
• T\begin{align*}T\end{align*} or FT\begin{align*}F_T\end{align*} ; the force of tension is a force that acts in strings, wires, ropes, and other non-stretchable lines of material.2\begin{align*}^2\end{align*}
• fμFN\begin{align*}f \le \mu F_N\end{align*} ; the force of friction acting between two surfaces, can be at rest (static friction, fs)\begin{align*}(\text{static friction,} \ f_s)\end{align*} or moving (kinetic friction,fk)2\begin{align*}(\text{kinetic friction}, f_k)^2\end{align*}
• P=FA\begin{align*}P = \frac{F}{A}\end{align*} ; Pressure is a force exerted over some area

Problem Solving for Newton’s Laws, Step-By-Step

1. Figure out which object is “of interest.”

If you're looking for the motion of a rolling cart, the cart is the object of interest. Draw a sketch! This may help you sort out which object is which in your problem.

2. Using your object as the “origin”, draw an xy\begin{align*}x-y\end{align*} coordinate system on your sketch.

This will help you properly place the directions of all your forces. Label to the right the +x\begin{align*}+x\end{align*} direction and up as the +y\begin{align*}+y\end{align*} direction. Label the mass.

3. Identify all the forces acting on the object and draw them. This is called a free body diagram (FBD). LABEL all forces – not with numbers, but with the symbol representing the force; i.e. T\begin{align*}T\end{align*} is tension, mg is weight, N\begin{align*}N\end{align*} is normal, etc. Be careful - If you can’t identify a force it may not really exist! INERTIA IS NOT A FORCE!

a. If the object has mass and is near the Earth, the easiest (and therefore first) force to write down is the force of gravity, pointing downward, with value mg.

b. If the object is in contact with a surface, it means there is a normal force acting on the object. This normal force points away from and is perpendicular to the surface.

c. There may be more than one normal force acting on an object. For instance, if you have a bologna sandwich, remember that the slice of bologna feels normal forces from both the slices of bread!

d. If a rope, wire, or cord is pulling on the object in question, you've found yourself a tension force. The direction of this force is in the same direction that the rope is pulling (you can’t push on a rope!). Don’t worry about what’s on the OTHER end of the rope – it’s just “tension”.

e. Remember that Newton's 3rd Law, calling for “equal and opposite forces,” does not apply to a single object. Only include forces acting on the ONE object you have identified.

f. Recall that scales (like a bathroom scale you weigh yourself on) read out the normal force acting on you, not your weight. If you are at rest on the scale, the normal force equals your weight. If you are accelerating up or down, the normal force had better be higher or lower than your weight, or you won’t have an unbalanced force to accelerate you.

g. Never include “ma” as a force acting on an object. “ma” is the result for which the net force Fnet\begin{align*}F_{net}\end{align*} is the cause.

4. Identify which forces are in the x\begin{align*}x-\end{align*}direction, which are in the y\begin{align*}y-\end{align*}direction, and which are at an angle.

a. If a force is upward, make it in the y\begin{align*}y-\end{align*}direction and give it a positive sign. If it is downward, make it in the y\begin{align*}y-\end{align*}direction and give it a negative sign.

b. Same thing applies for right vs. left in the x\begin{align*}x-\end{align*}direction. Make rightward forces positive.

5. “Fill in” to Newton’s second law:

Findividual forces =ma.or  Fxdirection forces  =max Add forces from diagramalways ma on right side!\begin{align*}& \quad \qquad \ \sum F_{\text{individual forces}}\qquad \ = ma. \\ & \qquad \text{or} \ \ \sum F_{x-\text{direction forces}}\quad \ \ = ma_x \\ & \qquad \qquad \nearrow \qquad \qquad \qquad \qquad \qquad \ \nwarrow \\ & Add \ forces \ from \ diagram \qquad always \ ma \ on \ right \ side!\end{align*}

a. Remember that all the rightward forces add with a plus (+) sign, and that all the leftward forces add with a minus (–) sign.

b. Now repeat, but for the \begin{align*}y-\end{align*}forces and this will be equal to the mass multiplied by the acceleration in the \begin{align*}y-\end{align*}direction.

Solved Examples

For Newton Law Problems, in addition to the ‘5-Step Process’, ALWAYS draw a Free-Body Diagram (FBD), before attempting to answer the question or solve the problem.

Example 1: A 175-g bluebird slams into a window with a force of 190 N. What is the bird’s acceleration?

Question: \begin{align*}a = ? [m/s^2]\end{align*}

Given: \begin{align*}m = 175\ grams = 0.175\ kg\end{align*}

\begin{align*}{\;}\qquad \quad F = 19.0 \ N\end{align*}

Equation: \begin{align*}a = \frac{F_{net}}{m}\end{align*}

Plug n’ Chug: \begin{align*}a = \frac{F_{net}}{m} = \frac{19.0 \ N}{0.175 \ kg} = \frac{19.0 \frac{kg \cdot m}{s^2}}{0.175\ kg} = 109 \frac{m}{s^2}\end{align*}

Answer: \begin{align*}\boxed{\mathbf{109 \ m/s^2}}\end{align*}

Example 2: Calculate the acceleration of a rocket that has 500N of thrust force and a mass of 10kg.

Question: \begin{align*}a = ? [m/s^2]\end{align*}

Given: \begin{align*}m = 10\ kg\end{align*}

\begin{align*}{\;} \qquad \quad F_{\text{thrust}} = 500\ N\end{align*}

\begin{align*}{\;} \qquad \quad g = 10.0\ m/s^2\end{align*}

Equations: \begin{align*}\sum F_{\text{individual forces}} = ma\end{align*}

or, in this case, \begin{align*}\sum F_{y-\text{direction forces}} = ma_y\end{align*}

Plug nChug: Use FBD to “fill in” Newton’s second law equation:

\begin{align*}\sum F_{y-\text{direction forces}} &= ma_y \\ F - Mg & = Ma \\ 500N - 10\ kg(10\ m/s^2) & = 10kg (a) \\ a & = 40\ m/s^2\end{align*}

Example 3: Calculate the force necessary to slide a 4.7-kg chair across a room at a constant speed if the coefficient of kinetic friction between the chair and the floor is 0.68.

Question: \begin{align*}F = ? [N]\end{align*}

Given: \begin{align*}m = 4.7\ kg\end{align*}

\begin{align*}{\;}\qquad \quad \mu_k = 0.68\end{align*}

\begin{align*}{\;} \qquad \quad g = 10.0\ m/s^2\end{align*}

Equations: \begin{align*}\sum F = ma\end{align*}

\begin{align*}{\;}\qquad \ \qquad \sum F_y = N - mg = 0, \text{so}\ N = mg\end{align*}

\begin{align*}{\;}\qquad \ \qquad \sum F_x = ma_x\end{align*}

\begin{align*}{\;}\qquad \ \qquad F_{\text{pull}} - f_k = 0 \ (\text{because the chair is moving at}\ constant \ speed,\ \text{so}\ a=0)\end{align*}

\begin{align*}{\;}\qquad \ \qquad F_{\text{pull}} = \mu_k N\end{align*}

Plug n’ Chug: The force necessary to move the chair at a constant speed is equal to the frictional force between the chair and the floor. However in order to calculate the frictional force you must first determine the normal force which is (in this case) equal to the weight (i.e. \begin{align*}F_g\end{align*}) of the chair.

\begin{align*}N & = mg = (4.7 \ kg)(10 \ m/s^2) = 47 \ N \\ F_{\text{pull}} &= \mu_k \ N = (0.68)(47 \ N) = 32 \ N\end{align*}

Answer: \begin{align*}\boxed{\mathbf{32 \ N}}\end{align*}

Example 4: How much pressure does a 340-g Coke can exert on a table if the diameter of the can is 8.0 cm?

Question: \begin{align*}P = ? [Pa]\end{align*}

Given: \begin{align*}m = 340\ grams = 0.340\ kg\end{align*}

\begin{align*}{\;}\qquad \quad \text{diameter} = 8.0\ cm = 0.08\ m\end{align*}

\begin{align*}{\;}\qquad \quad g = 10\ m/s^2\end{align*}

Equations: \begin{align*}P = \frac{F}{A}\end{align*} Area of a circle: \begin{align*}A = \pi \cdot r^2\end{align*}

Plug n’ Chug: \begin{align*}P = \frac{F}{A} = \frac{mg}{ \pi \cdot r^2} = \frac{(0.340 \ kg)\left (10 \frac{m}{s^2} \right )}{\pi \left (\frac{0.08 \ m}{2} \right )^2}= \left (\frac{3.4 \frac{kg \cdot m}{s^2}}{5.0 \times 10^{-3} m^2} \right ) = 680 \frac{N}{m^2} = 680 \ Pa\end{align*}

Answer: \begin{align*}\boxed{\mathbf{680 \ Pa}}\end{align*}

Newton’s Laws Problem Set

1. Is there a net force on a hammer when you hold it steady above the ground? If you let the hammer drop, what’s the net force on the hammer while it is falling to the ground?
2. If an object is moving at constant velocity or at rest, what is the minimum number of forces acting on it (other than zero)?
3. If an object is accelerating, what is the minimum number of forces acting on it?
4. You are standing on a bathroom scale. Can you reduce your weight by pulling up on your shoes? (Try it.) Explain.
5. When pulling a paper towel from a paper towel roll, why is a quick jerk more effective than a slow pull?
6. If you hang from the monkey bars with your arms completely vertical, what’s the force in each of your arms compared to your weight? Does that force increase, decrease, or remain constant if you reach over several bars with one arm, thus increasing the angle between your arms? Is it possible to hand with your arms completely horizontal? Why not? Explain in terms of force components; draw FBDs to support your argument.
7. You’re riding in a train moving at a constant velocity and you flip a coin. When it hits the floor, where does it land compared to where you dropped it? If instead the coin lands a meter away, what will you assume about the motion of the train? Explain using Newton’s law of inertia.
8. When hit from behind in a car crash, a passenger can suffer a neck injury called whiplash. Explain in terms of inertia how this occurs, and how headrests can prevent the injury.
9. A cheetah can outrun a gazelle in a short straight race, but the gazelle can escape with its life by zigzagging. The cheetah is more massive than the gazelle – explain how this strategy works.
10. If your hammer develops a loose head, you can tighten it by banging it on the ground. A little physics secret though – it’s better to bang the hammer head up rather than head down. Explain, using inertia.
11. A car moves down the road with a constant positive velocity. A negative force is applied. Explain how this could occur, and what happens to the speed of the car as a result.
12. During a rocket launch, the rocket’s acceleration increases greatly over time. Explain, using Newton’s Second Law. (Hint: most of the mass of a rocket on the launch pad is fuel).
13. Why does a sharp knife cut so much better than a dull one?
14. Go online and find out what is meant by saying a watch has “jeweled bearings”. What’s the purpose of those bearings, and why are they made of jewels?
15. As a sky diver free falls from a plane what happens to the force of air resistance acting on her? What happens to the NET force acting on her? Explain what is meant by “terminal velocity” and why it occurs, in terms of Newton’s second law.
16. A stone with a mass of 10 kg is sitting on the ground, not moving.
1. What is the weight of the stone?
2. What is the normal force acting on the stone?
17. For a boy who weighs 500 N on Earth what are his mass and weight on the moon (where \begin{align*}g = 1.6 \ m/s^2\end{align*})?
18. The man is hanging from a rope wrapped around a pulley and attached to both of his shoulders. The pulley is fixed to the wall. The rope is designed to hold 500 N of weight; at higher tension, it will break. Let’s say he has a mass of 80 kg. Draw a free body diagram and explain (using Newton’s Laws) whether or not the rope will break.
19. Now the man ties one end of the rope to the ground and is held up by the other. Does the rope break in this situation? What precisely is the difference between this problem and the one before?
20. Draw arrows representing the forces acting on the cannonball as it flies through the air. Assume that air resistance is small compared to gravity, but not negligible.
21. Draw free body diagrams (FBDs) for all of the following objects involved (in bold) and label all the forces appropriately. Make sure the lengths of the vectors in your FBDs are proportional to the strength of the force: smaller forces get shorter arrows! (a) A man stands in an elevator that is accelerating upward at \begin{align*}2 \ m/s^2\end{align*}. (b) A boy is dragging a sled at a constant speed. The boy is pulling the sled with a rope at a \begin{align*}30^\circ\end{align*} angle. (c) The picture shown here is attached to the ceiling by three wires. (d) A bowling ball rolls down a lane at a constant velocity. (e) A car accelerates down the road. There is friction \begin{align*}f\end{align*} between the tires and the road.
22. Mary is trying to make her 70.0-kg St. Bernard to out the back door but the dog refuses to walk. The force of friction between the dog and the floor is 350 N, and Mary pushes horizontally.
1. Draw a FBD for the problem after she gets the dog going. Label all forces.
2. If Mary pushes with a force of 400 N, what will be the dog’s acceleration?
3. If her dog starts from rest, what will be her dog’s final velocity if Mary pushes with a force of 400 N for 3.8 s?
4. How hard must Mary push in order to move the dog with a constant speed? Explain your answer.
23. A crane is lowering a box of mass 50 kg with an acceleration of \begin{align*}2.0 \ m/s^2\end{align*}.
1. Find the tension \begin{align*}F_T\end{align*} in the cable.
2. If the crane lowers the box at a constant speed, what is the tension \begin{align*}F_T\end{align*} in the cable?
24. A rocket of mass 10,000 kg is accelerating up from its launch pad. The rocket engines exert a vertical upward force of \begin{align*}3 \times 10^5 \ N\end{align*} on the rocket.
1. Calculate the weight of the rocket.
2. Draw a FBD for the rocket, labeling all forces.
3. Calculate the acceleration of the rocket (assuming the mass stays constant).
4. Calculate the height of the rocket after 12.6 s of acceleration, starting from rest.
5. In a REAL rocket, the mass decreases as fuel is burned. How would this affect the acceleration of the rocket? Explain briefly.
25. It’s a dirty little Menlo secret that every time the floors in Stent Hall are waxed, Mr. Colb likes to slide down the hallway in his socks. Mr. Colb weighs 950 N and the force of friction acting on him is 100 N.
1. Draw a FBD for Mr. Colb.
2. Calculate Mr. Colb’s acceleration down the hall.
3. Oh no! There’s an open door leading nowhere at the end of the second floor hallway! Mr. Colb is traveling at 2.8 m/s when he becomes a horizontally launched projectile and plummets to the ground below (don’t worry, he lands on a pile of backpacks and only his pride is injured). If the window is 3.7 m high, calculate how far from the base of the wall Mr. Colb lands.
26. A physics student weighing 500 N stands on a scale in an elevator and records the scale reading over time. The data are shown in the graph below. At time \begin{align*}t = 0\end{align*}, the elevator is at rest on the ground floor.
1. Draw a FBD for the person, labeling all forces.
2. What does the scale read when the elevator is at rest?
3. Calculate the acceleration of the person from 5-10 sec.
4. Calculate the acceleration of the person from 10-15 sec. Is the passenger at rest?
5. Calculate the acceleration of the person from 15-20 sec. In what direction is the passenger moving?
6. Is the elevator at rest at \begin{align*}t=25 \ s\end{align*}. Justify your answer.
27. Nathan pulls his little brother in a wagon, using a rope inclined at \begin{align*}30^\circ\end{align*} above the horizontal. The wagon and brother have a total mass of 80 kg, the average coefficient of friction between the wagon wheels and the floor is 0.08, and Nathan pulls on the rope with a force of 100 N.
1. Draw a force diagram for the wagon, labeling all forces.
2. Calculate the horizontal and vertical components of Nathan’s pull. Label them on your diagram (use dotted lines for components so as not to confuse them with other forces).
3. Calculate the normal force acting on the wagon. (HINT: It is NOT equal to the weight! Use your FBD above).
4. Calculate the force of friction on the wagon.
5. Calculate the horizontal acceleration of the wagon.
28. When the 20 kg box to the right is pulled with a force of 100 N, it just starts to move (i.e. the maximum value of static friction is overcome with a force of 100 N). What is the value of the coefficient of static friction, \begin{align*}\mu_S\end{align*}?
29. A different box, this time 5 kg in mass, is being pulled with a force of 20 N and is sliding with an acceleration of \begin{align*}2 \ m/s^2\end{align*}. Find the coefficient of kinetic friction, \begin{align*}\mu_K\end{align*}.
30. Every day Fakir likes to spend about an hour meditating on his bed of nails. Fakir’s mass is 60 kg, his bed contains 2000 nails, and each nail point has a surface area of \begin{align*}4 \ mm^2\end{align*}.
1. Calculate the total surface area of all the nail points, then convert that area into square meters.
2. Calculate the pressure exerted on Fakir by the nails. Compare this to your answer from the previous question.
31. Estimate the pressure in Pascals that you exert on the ground when you stand in your bare feet. Clearly state your assumptions.
32. A VW Bug hits a huge truck head-on. Each vehicle was initially going 50 MPH.
1. Which vehicle experiences the greater force?
2. Which experiences the greater acceleration? Explain briefly.
33. Is it possible for me to wave my hand and keep the rest of my body perfectly still? Why or why not?
34. How does a rocket accelerate in space, where there is nothing to ‘push off’ against?

Answers (using \begin{align*}g = 10 \ m/s^2\end{align*}):

1. Zero; weight of the hammer minus the air resistance.
2. 2 forces
3. 1 force
4. No
5. The towel’s inertia resists the acceleration
1. 100 N
2. 100 N
6. 50 kg; 80 N
7. The rope will not break because his weight of 800 N is distributed between the two ropes.
8. Yes, because his weight of 800 N is greater than what the rope can hold.
1. (b) \begin{align*}a = 0.71 \ m/s^2\end{align*} (c) \begin{align*}v = 2.7 \ m/s\end{align*} (d) 350 N
1. 400 N
2. 500 N
1. 100,000 N
2. \begin{align*}20 \ m/s^2\end{align*}
2. (b) \begin{align*}-1.1 \ m/s^2\end{align*} (c) 2.4 m
3. (b) 500 N (c) \begin{align*}6 \ m/s^2\end{align*} (d) 0 (e) \begin{align*}-4 \ m/s^2\end{align*}
4. (b) \begin{align*}F_x = 87 \ N, F_y = 50 \ N\end{align*} (c) \begin{align*}N = 750 \ N\end{align*} (d) \begin{align*}f = 60 \ N\end{align*} (e) \begin{align*}0.34 \ m/s^2\end{align*}
5. 0.5
6. 0.2
1. \begin{align*}0.008 \ m^2\end{align*}
2. 75 kPa

OPTIONAL PROBLEMS

1. The physics professor holds an eraser up against a wall by pushing it directly against the wall with a completely horizontal force of 20 N. The eraser has a mass of 0.5 kg. The wall has coefficients of friction \begin{align*}\mu_S = 0.8\end{align*} and \begin{align*}\mu_K = 0.6\end{align*}.
1. Draw a free body diagram for the eraser.
2. What is the normal force \begin{align*}F_N\end{align*} acting on the eraser?
3. What is the maximum mass \begin{align*}m\end{align*} the eraser could have and still not fall down?
4. What would happen if the wall and eraser were both frictionless?
2. A tug of war erupts between you and your sweetie. Assume your mass is 60 kg and the coefficient of friction between your feet and the ground is 0.5 (good shoes). Your sweetie’s mass is 85 kg and the coefficient of friction between his/her feet and the ground is 0.35 (socks). Who is going to win? Explain, making use of a calculation.
3. A stunt driver is approaching a cliff at very high speed. Sensors in his car have measured the acceleration and velocity of the car, as well as all forces acting on it, for various times. The driver’s motion can be broken down into the following steps: Step 1: The driver, beginning at rest, accelerates his car on a horizontal road for ten seconds. Sensors show that there is a force in the direction of motion of 6000 N, but additional forces acting in the opposite direction with magnitude 1000 N. The mass of the car is 1250 kg. Step 2: Approaching the cliff, the driver takes his foot of the gas pedal (There is no further force in the direction of motion.) and brakes, increasing the force opposing motion from 1000 N to 2500 N. This continues for five seconds until he reaches the cliff. Step 3: The driver flies off the cliff, which is 44.1 m high, and begins projectile motion. (a) Ignoring air resistance, how long is the stunt driver in the air? (b) For Step 1: i. Draw a free body diagram, naming all the forces on the car. ii. Calculate the magnitude of the net force. iii. Find the change in velocity over the stated time period. iv. Make a graph of velocity in the \begin{align*}x-\end{align*}direction vs. time over the stated time period. v. Calculate the distance the driver covered in the stated time period. Do this by finding the area under the curve in your graph of (iv). Then, check your result by using the equations for kinematics. (c) Repeat (b) for Step 2. i. Draw a free body diagram, naming all the forces on the car. ii. Calculate the magnitude of the net force. iii. Find the change in velocity over the stated time period. iv. Make a graph of velocity in the \begin{align*}x-\end{align*}direction vs. time over the stated time period. v. Calculate the distance the driver covered in the stated time period. Do this by finding the area under the curve in your graph of (iv). Then, check your result by using the equations for kinematics. (d) Calculate the distance that the stunt driver should land from the bottom of the cliff.
4. The large box on the table is 30 kg and is connected via a rope and pulley to a smaller 10 kg box, which is hanging. The 10 kg mass is the highest mass you can hang without moving the box on the table. Find the coefficient of static friction \begin{align*}\mu_S\end{align*}.

1. (b) 20 N (c) 4.9 N (d) Eraser would slip down the wall
2. You will win because you have a slightly larger frictional force (300N compared to 297.5N)
3. (a) 3 seconds (d) 90 m
4. \begin{align*}\mu_s = \frac{1}{3}\end{align*}

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