## 5 Steps to a 5: AP Physics C (2016)

### STEP __4__

### Review the Knowledge You Need to Score High

### CHAPTER 12

### Newton”s Second Law, F_{net} = *ma*

**IN THIS CHAPTER**

**Summary:** __Chapter 10__ explained how to deal with objects in equilibrium, that is, with zero acceleration. The same problem-solving process can be used with accelerating objects.

**Key Ideas**

Only a NET force, not an individual force, can be set equal to *ma* .

Use a free-body diagram and the four-step problem-solving process when a problem involves forces.

When two masses are connected by a rope, the rope has the same tension throughout. (One rope = one tension.)^{ }^{1}

Newton”s third law: force pairs must act on different objects.

**Relevant Equations**

Um, the chapter title says it all …

What this means is that *the net force acting on an object is equal to the mass of that object multiplied by the object”s acceleration* . And that statement will help you with all sorts of problems.

**The Four-Step Problem-Solving Process**

If you decide that the best way to approach a problem is to use *F* _{net} = *ma* , then you should solve the problem by following these four steps.

- Draw a proper free-body diagram.
- Resolve vectors into their components.
- For each axis, set up an expression for
*F*_{net}, and set it equal to*ma*. - Solve your system of equations.

Note the marked similarity of this method to that discussed in the chapter on equilibrium.

Following these steps will get you majority credit on an AP free-response problem even if you do not ultimately get the correct answer. In fact, even if you only get through the first one or two steps, it is likely that you will still get some credit.

**Only Net Force Equals ma**

**THIS IS REALLY IMPORTANT.** Only *F* _{net} can be set equal to *ma* . You cannot set any old force equal to *ma* . For example, let”s say that you have a block of mass *m* sitting on a table. The force of gravity, *mg* , acts down on the block. But that does not mean that you can say, “*F* = *mg* , so the acceleration of the block is *g* , or about 10 m/s^{2} .” Because we know that the block isn”t falling! Instead, we know that the table exerts a normal force on the block that is equal in magnitude but opposite in direction to the force exerted by gravity. So the NET force acting on the block is 0. You can say “*F* _{net} = 0, so the block is not accelerating.”

**A Simple Example**

A block of mass *m* = 2 kg is pushed along a frictionless surface. The force pushing the block has a magnitude of 5 N and is directed at *θ* = 30° below the horizontal. What is the block”s acceleration?

We follow our four-step process. First, draw a proper free-body diagram.

Second, we break the *F* _{push} vector into components that line up with the horizontal and vertical axes.

We can now move on to __Step 3__ , writing equations for the net force in each direction:

Now we can plug in our known values into these equations and solve for the acceleration of the block. First, we solve for the right–left direction:

Next, we solve for the up–down direction. Notice that the block is in equilibrium in this direction—it is neither flying off the table nor being pushed through it—so we know that the net force in this direction must equal 0.

So the acceleration of the block is simply 2.2 m/s^{2} to the right.

*F* _{net} on Inclines

A block of mass *m* is placed on a plane inclined at an angle *θ* . The coefficient of friction between the block and the plane is *μ* . What is the acceleration of the block down the plane?

This is a really boring problem. But it”s also a really common problem, so it”s worth looking at.^{ }^{2}

Note that no numbers are given, just variables. That”s okay. It just means that our answer should be in variables. Only the given variables—in this case *m, θ* , and *m* —and constants such as *g* can be used in the solution. And you shouldn”t plug in any numbers (such as 10 m/s^{2} for *g* ), even if you know them.

* Step 1 :* Free-body diagram.

* Step 2 :* Break vectors into components.

If you don”t know where we got those vector components, refer back to __Chapter 9__ .

* Step 3 :* Write equations for the net force in each direction.

Note that the block is in equilibrium in the direction perpendicular to the plane, so the equation for *F* _{net, perpendicular} (but not the equation for *F* _{net, down the plane} ) can be set equal to 0.

* Step 4 :* Solve.

We can rewrite *F* _{f} , because

*F _{f} = μF_{N} = μ mg* (cos

*θ*)

Plugging this expression for *F* _{f} into the “*F* _{net, down the plane} ” equation, we have

It always pays to check the reasonability of the answer. First, the answer doesn”t include any variables that weren”t given. Next, the units work out: *g* has acceleration units; neither the sine or cosine of an angle nor the coefficient of friction has any units.

Second, compare the answer to something familiar. Note that if the plane were vertical, *θ* = 90°, so the acceleration would be *g* —yes, the block would then be in free fall! Also, note that friction tends to make the acceleration smaller, as you might expect.

For this particular incline, what coefficient of friction would cause the block to slide with constant speed?

Constant speed means *a* = 0. The solution for *F _{N} *in the perpendicular direction is the same as before:

*F*

_{N}_{ }=

*mg*(cos

*θ*). But in the down-the-plane direction, no acceleration means that

*F*

_{f}_{ }=

*mg*(sin

*θ*). Because

*μ*=

*F*

_{f}_{ }/

*F*

_{N}_{ },

Canceling terms and remembering that sin/cos = tan, you find that *μ* = tan*θ* when acceleration is zero.

You might note that neither this answer nor the previous one includes the mass of the block, so on the same plane, both heavy and light masses move the same way!

*F* _{net} for a Pulley

Before we present our next practice problem, a few words about tension and pulleys are in order. Tension in a rope is the same *everywhere* in the rope, even if the rope changes direction (such as when it goes around a pulley) or if the tension acts in different directions on different objects. ONE ROPE = ONE TENSION. If there are multiple ropes in a problem, each rope will have its own tension. TWO ROPES = TWO TENSIONS.^{ }^{3}

When masses are attached to a pulley, the pulley can only rotate one of two ways. Call one way positive, the other, negative.

A block of mass *M* and a block of mass *m* are connected by a thin string that passes over a light frictionless pulley. Find the acceleration of the system.

We arbitrarily call counterclockwise rotation of the pulley “positive.”

* Step 1 :* Free-body diagrams.

The tension T is the same for each block—ONE ROPE = ONE TENSION. Also, note that because the blocks are connected, they will have the same acceleration, which we call *a* .

* Step 2 :* Components.

The vectors already line up with one another. On to __Step 3__ .

* Step 3 :* Equations.

Notice how we have been careful to adhere to our convention of which forces act in the positive and negative directions.

* Step 4 :* Solve.

Let”s solve for *T* using the first equation:

*T* = *Mg* − *Ma* .

Plugging this value for *T* into the second equation, we have

(*Mg* − *Ma* ) − *mg* = *ma*

Our answer is

A 2-kg block and a 5-kg block are connected as shown on a frictionless surface. Find the tension in the rope connecting the two blocks. Ignore any friction effects.

Why don”t you work this one out for yourself? We have included our solution on the following page.

**Solution to Example Problem**

* Step 1 :* Free-body diagrams.

* Step 2 :* Components.

Again, our vectors line up nicely, so on to __Step 3__ .

* Step 3 :* Equations.

Before we write any equations, we must be careful about signs: we shall call counterclockwise rotation of the pulley “positive.”

For the more massive block, we know that, because it is not flying off the table or tunneling into it, it is in equilibrium in the up–down direction. But it is not in equilibrium in the right–left direction.

For the less massive block, we only have one direction to concern ourselves with: the up–down direction.

*F* _{net} = *T* − *mg* = *ma*

We can solve for *T* from the “*F* _{net,} * _{x} *” equation for the more massive block and plug that value into the “

*F*

_{net}” equation for the less massive block, giving us

(−*Ma* ) − *mg* = *ma*

We rearrange some terms to get

Now we plug in the known values for *M* and *m* to find that

To finish the problem, we plug in this value for *a* into the “*F* _{net,} * _{x} *” equation for the more massive block.

**More Thoughts on F _{net} = ma**

The four example problems in this chapter were all solved using only *F* _{net} = *ma* . Problems you might face in the real world—that is, on the AP test—will not always be so straightforward. Here”s an example: imagine that this last example problem asked you to find the speed of the blocks after 2 seconds had elapsed, assuming that the blocks were released from rest. That”s a kinematics problem, but to solve it, you have to know the acceleration of the blocks. You would first have to use *F* _{net} = *ma* to find the acceleration, and then you could use a kinematics equation to find the final speed. We suggest that you try to solve this problem: it”s good practice.

Also, remember in __Chapter 12__ when we introduced the unit of force, the newton, and we said that 1 N = 1 kg·m/s^{2} ? Well, now you know why that conversion works: the units of force must be equal to the units of mass multiplied by the units of acceleration.

**Exam tip from an AP Physics veteran:**

Newton”s second law works for *all* kinds of forces, not just tensions, friction, and such. Often what looks like a complicated problem with electricity or magnetism is really just an *F* _{net} = *ma* problem, but the forces might be electric or magnetic in nature.

*—Jonas, high school senior*

**Newton”s Third Law**

We”re sure you”ve been able to quote the third law since birth, or at least since 5th grade: “Forces come in equal and opposite action-reaction pairs,” also known as “For every action there is an equal and opposite reaction.” If I push down on the Earth, the Earth pushes up on me; a football player who makes a tackle experiences the same force that he dishes out.

What”s so hard about that? Well, ask yourself one of the most important conceptual questions in first-year physics: “If all forces cause reaction forces, then how can anything ever accelerate?” Pull a little lab cart horizontally across the table … you pull on the cart, the cart pulls on you, so don”t these forces cancel out, prohibiting acceleration?

Well, obviously, things can move. The trick is, Newton”s third law force pairs *must act on different objects* , and so can never cancel each other.

When writing *F* _{net} = *ma* , only consider the forces acting on the object in question. Do not include forces exerted *by* the object.

Consider the lab cart. The only horizontal force that it experiences is the force of your pull. So, it accelerates toward you. Now, you experience a force from the cart, but you also experience a whole bunch of other forces that keep you in equilibrium; thus, you don”t go flying into the cart.

**This Chapter Was Not as Easy as You Thought**

Be careful with this chapter. Most Physics C students say, “Oh, come on, this stuff is easy … let”s move on to something challenging.” Okay, you”re right—if you”re at the level you need to be for Physics C, basic Newton”s second law problems need to be easy for you. What you must remember from this and the equilibrium chapter is the **absolute necessity** of free-body diagrams.

No matter how easy or hard an *F* _{net} problem may seem, you must start the problem with a free-body diagram. Points are awarded for the free-body diagram, and that diagram will ensure that you don”t make minor mistakes on the rest of the problem. My own Physics C students frequently mess up on what should be straightforward problems when they try to take shortcuts. If you draw the FBD and follow the four-step problem-solving procedure, it”s hard to go wrong. Even professional physicists use free-body diagrams. You must, too.

** Practice Problems**

**Multiple Choice:**

** 1 .** A 2.0-kg cart is given a shove up a long, smooth 30° incline. If the cart is traveling 8.0 m/s after the shove, how much time elapses until the cart returns to its initial position?

(A) 1.6 s

(B) 3.2 s

(C) 4.0 s

(D) 6.0 s

(E) 8.0 s

** 2 .** A car slides up a frictionless inclined plane. How does the normal force of the incline on the car compare with the weight of the car?

(A) The normal force must be equal to the car”s weight.

(B) The normal force must be less than the car”s weight.

(C) The normal force must be greater than the car”s weight.

(D) The normal force must be zero.

(E) The normal force could have any value relative to the car”s weight.

** 3 .** In the diagram above, a 1.0-kg cart and a 2.0-kg cart are connected by a rope. The spring scale reads 10 N. What is the tension in the rope connecting the two carts? Neglect any friction.

(A) 30 N

(B) 10 N

(C) 6.7 N

(D) 5.0 N

(E) 3.3 N

** 4 .** The velocity–time graph above represents the motion of a 5-kg box. The only force applied to this box is a person pushing. Assuming that the box is moving to the right, what is the magnitude and direction of the force applied by the person pushing?

(A) 2.0 N, right

(B) 2.0 N, left

(C) 0.4 N, right

(D) 0.4 N, left

(E) 12.5 N, left

**Free Response:**

** 5 .** A 2-kg block and a 5-kg block are connected as shown above. The coefficient of friction between the 5-kg block and the flat surface is

*θ*= 0.2.

(A) Calculate the magnitude of the acceleration of the 5-kg block.

(B) Calculate the tension in the rope connecting the two blocks.

** 6 .** Bert, Ernie, and Oscar are discussing the gas mileage of cars. Specifically, they are wondering whether a car gets better mileage on a city street or on a freeway. All agree (correctly) that the gas mileage of a car depends on the force that is produced by the car”s engine—the car gets fewer miles per gallon if the engine must produce more force. Whose explanation is completely correct?

**Bert says:** Gas mileage is better on the freeway. In town the car is always speeding up and slowing down because of the traffic lights, so because *F* _{net} = *ma* and acceleration is large, the engine must produce a lot of force. However, on the freeway, the car moves with constant velocity, and acceleration is zero. So the engine produces no force, allowing for better gas mileage.

**Ernie says:** Gas mileage is better in town. In town, the speed of the car is slower than the speed on the freeway. Acceleration is velocity divided by time, so the acceleration in town is smaller. Because *F* _{net} = *ma* , then, the force of the engine is smaller in town giving better gas mileage.

**Oscar says:** Gas mileage is better on the freeway. The force of the engine only has to be enough to equal the force of air resistance—the engine doesn”t have to accelerate the car because the car maintains a constant speed. Whereas in town, the force of the engine must often be greater than the force of friction and air resistance in order to let the car speed up.

** Solutions to Practice Problems**

** 1 . B—** “Smooth” usually means, “ignore friction.” So the only force acting along the plane is a component of gravity,

*mg*(sin 30°). The

*F*

_{net}equation becomes

*mg*(sin 30°) − 0 =

*ma*. The mass cancels, leaving the acceleration as 5 m/s

^{2}. What”s left is a kinematics problem. Set up a chart, calling the direction down the plane as positive:

Use ** (Δ*x* = *v* _{0} *t* + ^{1} /_{2} *at* ^{2} ) to find that the time is 3.2 s.

** 2 . B—** The normal force exerted on an object on an inclined plane equals

*mg*(cos

*θ*), where

*θ*is the angle of the incline. If

*θ*is greater than 0, then cos

*θ*is less than 1, so the normal force is less than the object”s weight.

** 3 . E—** Consider the forces acting on each block separately. On the 1.0-kg block, only the tension acts, so

*T*= (1.0 kg)

*a*. On the 2.0-kg block, the tension acts left, but the 10 N force acts right, so 10 N −

*T*= (2.0 kg)

*a*. Add these equations together (noting that the tension in the rope is the same in both equations), getting 10 N = (3.0

*kg*)

*a*; acceleration is 3.3 m/s

^{2}. To finish,

*T*= (1.0 kg)

*a*, so tension is 3.3 N.

** 4 . B—** The acceleration is given by the slope of the

*v*–

*t*graph, which has magnitude 0.4 m/s

^{2}.

*F*

_{net}=

*ma*, so 5 kg × 0.4 m/s

^{2}= 2.0 N. This force is to the left because acceleration is negative (the slope is negative), and negative was defined as left.

** 5 .** The setup is the same as in the chapter”s example problem, except this time there is a force of friction acting to the left on the 5-kg block. Because this block is in equilibrium vertically, the normal force is equal to the block”s weight, 50 N. The friction force is

*μF*

_{N}_{ }, or 10 N.

Calling the down-and-right direction positive, we can write two equations, one for each block:

(A) To solve for acceleration, just add the two equations together. The tensions cancel. We find the acceleration to be 1.4 m/s^{2} .

(B) Plug back into either equation to find the final answer, that the tension is 17 N. This is more than the 14 N we found for the frictionless situation, and so makes sense. We expect that it will take more force in the rope to overcome friction on the table.

** 6 .** Although Bert is right that acceleration is zero on the freeway, this means that the NET force is zero; the engine still must produce a force to counteract air resistance. This is what Oscar says, so his answer is correct. Ernie”s answer is way off—acceleration is not velocity/time, acceleration is a CHANGE in velocity over time.

** Rapid Review**

- The net force on an object equals the mass of the object multiplied by the object”s acceleration.
- To solve a problem using
*F*_{net}=*ma*, start by drawing a good free-body diagram. Resolve forces into vector components. For each axis, the vector sum of forces along that axis equals*ma*_{i}, where*a*_{i}is the acceleration of the object along that axis. - When an object is on an inclined plane, resolve its weight into vector components that point parallel and perpendicular to the plane.
- For problems that involve a massless pulley, remember that if there”s one rope, there”s one tension.

^{1}^{ }But if the rope is attached across a massive pulley, the tension is different on each side of the pulley. See __Chapter 16__ .

^{2}^{ }If you want to make this problem more interesting, just replace the word “block” with the phrase “maniacal tobogganist” and the word “plane” with the phrase “highway on-ramp.”

^{3}^{ }Except for the physics C corollary, when the pulley is massive—this situation is discussed in __Chapter 16__ .

**CHAPTER 12**

**Forces**

** 1 .** A light string holds three identical masses over a lightweight pulley as shown in the figure. When the masses are released, the acceleration of the masses will be

(A) 2g

(B) g

(C) 2g/3

(D) g/2

(E) g/3

** 2 .** While removing paint, a carpenter accelerates a rough sanding block up a wall as shown. The coefficient of friction between the block and the wall is μ. What is the net force on the sanding block?

(A) *F* + *mg* + *F _{f}*

(B) *F* cos θ - *mg* - μ*F* sin θ

(C) *F* sin θ - *mg* - μ*F* cos θ

(D) *F* sin θ + *F _{N}*

(E) *F* + *mg* + *F _{N}*

** 3 .** The net force applied to an object as a function of time is shown in the figure. Which of the following position-time and velocity-time graphs are consistent with the force-time graph?

(A)

(B)

(C)

(D)

(E)

** 4 .** A force (

*F*) accelerates three blocks across a rough horizontal surface as shown in the figure. The coefficient of friction (μ) between the blocks and the surface is the same for each block. Which of the following statements is correct?

(A) The acceleration of all the blocks will be equal to .

(B) The net force acting on each block will be the same.

(C) The net force acting on block *m* is the largest.

(D) The net force acting on block 2*m* is the largest.

(E) The net force acting on block 3*m* is the largest.

** Answers**

__1__ .**E** —

__2__ .**B** —The horizontal forces acting on the block cancel each other out: *Fsin* θ=*F _{N} *. The vertical forces accelerate the block upward:

__3__ .**A** —The net force is a constant negative value, and the acceleration must also be constant and negative. Therefore, the velocity-time graph must have a negative slope. This eliminates choice B. Choices C and D start with a velocity of zero, yet their position-time graphs both begin with a positive slope, which is inconsistent. Choice E ends with a zero velocity, yet the position-time graph ends with a negative slope, which is inconsistent. This leaves choice A as the only viable graph that matches the force-time graph.

__4__ .**E** —All the blocks move as a unit, with the same acceleration of . Since each block has the same acceleration, the block with the largest mass will require the largest net force acting on it to maintain the same acceleration.