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AP Physics 1 : Newton's Second Law

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Example Questions

Example Question #51 : Newton's Second Law

A 32N force is applied to a block causing it to accelerate at $8\frac{m}{s^2}$ across concrete. The friction acts with 12N of force. What is the mass of the block?

Possible Answers:

m=0.8kg

m=1.2kg

m=5.7kg

m=2.5kg

Correct answer:

m=2.5kg

Explanation:

We begin by drawing a free body diagram of the block.

Force pushing on a block w friction

is the force applied to the block.

is the weight of the block, or the force due to gravity. Weight is defined as W=mg where is the mass of the block and is the gravitational constant.

F_N is the normal force acting perpendicular to the contact surface.

is the force due to kinetic friction.

Newton's second law is $F_{net}=ma$ where $F_{net}$ is the net force, is the mass of the block and is the acceleration.

We now apply Newton's second law in the direction of acceleration. In this problem, that is the x-direction.

Solving for mass, we have

$m=\frac{F_{net}}{a}$

There are two forces in the direction of acceleration,the applied force and the force due to friction . Assuming that is applied in the direction of acceleration and is in the opposite direction,

$F_{net}=F-f=32N-12N=20N$ . The problem tells us $a=8\frac{m}{s^2}$ . Substituting in this information gives us

$m=\frac{F_{net}}{a}=\frac{20}{8}=2.5kg$

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Example Question #51 : Newton's Second Law

A 64N force is applied to a block causing it to accelerate at $4\frac{m}{s^2}$ across wood. The friction acts with 36N of force. What is the mass of the block?

Possible Answers:

m=5kg

m=9kg

m=7kg

m=3kg

Correct answer:

m=7kg

Explanation:

We begin by drawing a free body diagram of the block.

Force pushing on a block w friction

is the force applied to the block.

is the weight of the block, or the force due to gravity. Weight is defined as W=mg where is the mass of the block and is the gravitational constant.

F_N is the normal force acting perpendicular to the contact surface.

is the force due to kinetic friction.

Newton's second law is $F_{net}=ma$ where $F_{net}$ is the net force, is the mass of the block and is the acceleration.

We now apply Newton's second law in the direction of acceleration. In this problem, that is the x-direction.

Solving for mass, we have

$m=\frac{F_{net}}{a}$

There are two forces in the direction of acceleration,the applied force and the force due to friction . Assuming that is applied in the direction of acceleration and is in the opposite direction,

$F_{net}=F-f=64N-36N=28N$ . The problem tells us $a=4\frac{m}{s^2}$ . Substituting in this information gives us

$m=\frac{F_{net}}{a}=\frac{28}{4}=7kg$

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Example Question #71 : Fundamentals Of Force And Newton's Laws

A 10kg block accelerates at $2\frac{m}{s^2}$ across concrete. The coefficient of friction of concrete is 0.5 . How much force was applied to the block?

Possible Answers:

F=35N

F=100N

F=70N

F=50N

Correct answer:

F=70N

Explanation:

We begin by drawing a free body diagram of the block.

Force pushing on a block w friction

is the force applied to the block.

is the weight of the block, or the force due to gravity. Weight is defined as W=mg where is the mass of the block and is the gravitational constant.

F_N is the normal force acting perpendicular to the contact surface.

is the force due to kinetic friction. Friction is defined as $f=\mu *F_N$ where $\mu$ is the coefficient of friction.

To find the force due to friction, we need to find F_N by applying Newton's second in the y-direction.

Newton's second law is $F_{net}=ma$ where $F_{net}$ is the net force, is the mass of the block and is the acceleration.

There are two forces in the y-direction, F_N and . There are in opposite directions, so they are subtracted. We are given m=10kg . There is no acceleration in the y-direction, so a=0 .

Substituting all this information into Newton's second law gives us

$F_{net}=ma\Rightarrow F_N-W=(10)*0\Rightarrow F_N-mg=(10)*0$

$F_N-mg=0\Rightarrow F_N=mg$

Assuming $g=10\frac{m}{s^2}$ ,

F_N=mg=(10)(10)=100N

Now that we have F_N , we can find the force due to friction. Given that $\mu=0.5$ and F_N=100N

$f=\mu *F_N=0.5*100N=50N$

We now apply Newton's second law in the direction of acceleration. In this problem, that is the x-direction.

There are two forces in the direction of acceleration,the applied force and the force due to friction . Assuming that is applied in the direction of acceleration and is in the opposite direction,

$F_{net}=F-f=F-50$ . The problem tells us m=10kg and $a=2\frac{m}{s^2}$ . Substituting this information into Newton's second law gives us

$F_{net}=ma$

F-50=10(2)

$F-50=20\Rightarrow F=70N$

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Example Question #71 : Fundamentals Of Force And Newton's Laws

A 8kg block accelerates at $3\frac{m}{s^2}$ across wood. The coefficient of friction wood is 0.2 . How much force was applied to the block?

Possible Answers:

F=40N

F=55N

F=20N

F=70N

Correct answer:

F=40N

Explanation:

We begin by drawing a free body diagram of the block.

Force pushing on a block w friction

is the force applied to the block.

is the weight of the block, or the force due to gravity. Weight is defined as W=mg where is the mass of the block and is the gravitational constant.

F_N is the normal force acting perpendicular to the contact surface.

is the force due to kinetic friction. Friction is defined as $f=\mu *F_N$ where $\mu$ is the coefficient of friction.

To find the force due to friction, we need to find F_N by applying Newton's second in the y-direction.

Newton's second law is $F_{net}=ma$ where $F_{net}$ is the net force, is the mass of the block and is the acceleration.

There are two forces in the y-direction, F_N and . There are in opposite directions, so they are subtracted. We are given m=8kg . There is no acceleration in the y-direction, so a=0 .

Substituting all this information into Newton's second law gives us

$F_{net}=ma\Rightarrow F_N-W=(8)*0\Rightarrow F_N-mg=(8)*0$

$F_N-mg=0\Rightarrow F_N=mg$

Assuming $g=10\frac{m}{s^2}$ ,

F_N=mg=(8)(10)=80N

Now that we have F_N , we can find the force due to friction. Given that $\mu=0.2$ and F_N=80N ,

$f=\mu *F_N=0.2*80N=16N$

We now apply Newton's second law in the direction of acceleration. In this problem, that is the x-direction.

There are two forces in the direction of acceleration, the applied force and the force due to friction . Assuming that is applied in the direction of acceleration and is in the opposite direction,

$F_{net}=F-f=F-16$ . The problem tells us m=8kg and $a=3\frac{m}{s^2}$ . Substituting this information into Newton's second law gives us

$F_{net}=ma$

F-16=8(3)

$F-16=24\Rightarrow F=40N$

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Example Question #294 : Forces

Suppose that a $2\: kg$ box sits on top of a $5\: kg$ box. A force is applied to the bottom box, causing both boxes to accelerate in the direction of the force. If the applied force remains the same, but the top box is removed, how does the acceleration of the system change?

Possible Answers:

The acceleration decreases by a factor of $\frac{7}{5}$

The acceleration does not change

The acceleration increases by a factor of $\frac{7}{5}$

The acceleration decreases by a factor of $\frac{5}{2}$

The acceleration increases by a factor of $\frac{5}{2}$

Correct answer:

The acceleration increases by a factor of $\frac{7}{5}$

Explanation:

To solve this question, we need to determine how the acceleration of the system changes once the top box is removed. To do this, we'll first have to evaluate the acceleration of the system before the top box is removed, and then do the same after its removal.

First, let's determine the initial acceleration. Since the applied force is causing both boxes to accelerate, we know that the total mass of the system has to be the combination of the masses of both boxes.

$F=(m_{bottom}+m_{top})a_{initial}$

Next, we can write an expression for the final acceleration of the system, which will only consist of the bottom box.

$F=m_{bottom}a_{final}$

Since the force applied doesn't change in either instance, we can relate each of the two expressions shown above.

$F=(m_{bottom}+m_{top})a_{initial}=m_{bottom}a_{final}$

Next, we can rearrange the above expression to isolate the term for final acceleration.

$a_{final}=a_{initial}(\frac{m_{bottom}+m_{top}}{m_{bottom}})$

Now, we just need to plug in the values that we know to obtain our answer.

$a_{final}=a_{initial}(\frac{5\: kg+2\: kg}{5\: kg})$

$a_{final}=\frac{7}{5}a_{initial}$

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Example Question #901 : Ap Physics 1

Force 1 pushes on a 5 kg ball with 15N to the right. Force 2 pushes on the ball with to the left. Force 3 pushes on the ball with to the left. The up and down forces can be assumed to be equivalent.

What is the acceleration of the ball?

Possible Answers:

$1.2 \frac{m}{s^2}$ to the left.

$20 \frac{m}{s^2}$ to the right.

$0.8 \frac{m}{s^2}$ to the right.

$0.8 \frac{m}{s^2}$ to the left.

$1.2 \frac{m}{s^2}$ to the right.

Correct answer:

$0.8 \frac{m}{s^2}$ to the right.

Explanation:

Acceleration can be found by dividing the sum of the forces acting on an object by the mass of the object:

$\vec{a}=\frac{\sum \vec{F}}{m}$

It is important to notice that both acceleration and the forces in the equation are vectors, meaning that both magnitude and direction matter.

In physics, we usually assign up and right as the positive directions, and down and left as the negative directions.

With that in mind, we first sum the forces:

$\sum F =15N-7N-4N=4N$

Because our result is positive, the sum of the forces is directed towards the right.

Plugging in our sum and given mass in to our acceleration equation we find:
$a=\frac{4N}{5kg}=0.8\frac{m}{s^2}$

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Example Question #902 : Ap Physics 1

An 80kg person is riding upward in an elevator. The elevator is coming to a stop, accelerating at $-5 \frac{m}{s^2}$ . What is the perceived weight of the person?

Possible Answers:

785N

423N

400N

193N

385N

Correct answer:

385N

Explanation:

The person's perceived weight is a result of gravity and the elevator's acceleration changing the normal force between person and elevator. Using Newton's second law, we can express all of the forces in the y-direction as:

$\Sigma F_{y} = F_{perceived} = N = mg + ma_{elevator}$ . Plugging in known values gives that $F_{perceived} = (80kg)(9.8 \frac{m}{s^2}) + (80kg)(-5 \frac{m}{s^2}) = 385N$

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Example Question #297 : Forces

A 2kg remote controlled helicopter is dropped from a height of 10m . The spinning rotors in the helicopter create a constant upward force of 10N . How long is the helicopter airborne before hitting the ground?

Possible Answers:

1.8s

1.4s

2.3s

2.0s

2.9s

Correct answer:

2.0s

Explanation:

The only forces acting on the helicopter are the upward mechanical force and gravity. Newton's second law gives that the net force can be given by $\Sigma F_{y} = F_{heli} + ma_{g} = ma_{tot}$ . The total acceleration can then be expressed as $a_{tot} = \frac{10N + (2kg)(-9.8 \frac{m}{s^2})}{2kg} = -4.8 \frac{m}{s^2}$ .

Using kinematics, we can find the time the helicopter is airborne. Using this kinematic equation $\Delta y = v_{0,y}t+\frac{1}{2}a_{y}t^2$ , we can plug in known values and solve for time:

$-10m = \frac{1}{2}(-4.8\frac{m}{s^2})t^2$ and t= 2.0s

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Example Question #51 : Newton's Second Law

A 100kg person slides down a ladder to the ground. If they started from rest, slid down at constant acceleration, and touched the ground later, what was the upward force on the person exerted by the ladder?

Possible Answers:

420N

395N

650N

850N

730N

Correct answer:

730N

Explanation:

The only forces acting on the person are gravity and an unidentified opposing force that slows the person's descent.

Using kinematics, we can find the net acceleration of the person:

$\Delta y = v_{0,y}t + \frac{1}{2}a_{y}t^2$ . Plugging in our known values, we find $-5m = \frac{1}{2}a_{y}(2s)^2$ and $a_{y} = -2.5 \frac{m}{s^2}$ .

Using Newton's second law, we know that

$\Sigma F_{y} = ma_{y}$ . All of the forces in the y-direction can then be expressed as $\Sigma F_{y} = F_{1} - mg = ma_{y}$ . Solving for $F_{1}$ we find that $F_{1} = (100kg)(-2.5\frac{m}{s^2}) + (100kg)(9.8\frac{m}{s^2}) = 730N$

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AP Physics 1 : Newton's Second Law

Study concepts, example questions & explanations for AP Physics 1

All AP Physics 1 Resources

Example Questions

Example Question #51 : Newton's Second Law

Example Question #51 : Newton's Second Law

Example Question #71 : Fundamentals Of Force And Newton's Laws

Example Question #71 : Fundamentals Of Force And Newton's Laws

Example Question #294 : Forces

Example Question #901 : Ap Physics 1

Example Question #902 : Ap Physics 1

Example Question #297 : Forces

Example Question #51 : Newton's Second Law

All AP Physics 1 Resources

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