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AP Physics 1 : Fundamentals of Force and Newton's Laws

Study concepts, example questions & explanations for AP Physics 1

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Example Questions

Example Question #281 : Forces

A 32N force is applied to a block causing it to accelerate at $8\frac{m}{s^2}$ across a sheet of ice. What is the mass of the block?

Possible Answers:

m=2kg

m=8kg

m=32kg

m=4kg

Correct answer:

m=4kg

Explanation:

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

Force pushing on a block

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.

Since the block is on ice, there is no friction.

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

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

Solving for mass, we have

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

The only force in the direction of acceleration is the applied force , therefore

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

$m=\frac{F_{net}}{a}=\frac{32}{8}=4kg$

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Example Question #281 : Forces

A 32N force is applied to a 10kg block sitting on a sheet of ice. What is the acceleration of the block?

Possible Answers:

$a=5\frac{m}{s^2}$

$a=3.2\frac{m}{s^2}$

$a=7.2\frac{m}{s^2}$

$a=1.8\frac{m}{s^2}$

Correct answer:

$a=3.2\frac{m}{s^2}$

Explanation:

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

Force pushing on a block

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.

Since the block is on ice, there is no friction.

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

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

Solving for acceleration, we have

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

The only force in the direction of acceleration is the applied force , therefore

$F_{net}=F=32N$ . The problem tells us m=10kg . Substituting in this information gives us

$a=\frac{F_{net}}{m}=\frac{32}{10}=3.2\frac{m}{s^2}$

Report an Error

Example Question #41 : Newton's Second Law

A 105N force is applied to a 45kg block sitting on a sheet of ice. What is the acceleration of the block?

Possible Answers:

$a=3.2\frac{m}{s^2}$

$a=5.13\frac{m}{s^2}$

$a=2.33\frac{m}{s^2}$

$a=1.52\frac{m}{s^2}$

Correct answer:

$a=2.33\frac{m}{s^2}$

Explanation:

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

Force pushing on a block

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.

Since the block is on ice, there is no friction.

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

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

Solving for acceleration, we have

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

The only force in the direction of acceleration is the applied force , therefore

$F_{net}=F=105N$ . The problem tells us m=45kg . Substituting in this information gives us

$a=\frac{F_{net}}{m}=\frac{105}{45}=2.33\frac{m}{s^2}$

Report an Error

Example Question #284 : Forces

A 64N force is applied to a block causing it to accelerate at $4\frac{m}{s^2}$ across a sheet of ice. What is the mass of the block?

Possible Answers:

m=32kg

m=8kg

m=4kg

m=16kg

Correct answer:

m=16kg

Explanation:

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

Force pushing on a block

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.

Since the block is on ice, there is no friction.

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

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

Solving for mass, we have

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

The only force in the direction of acceleration is the applied force , therefore

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

$m=\frac{F_{net}}{a}=\frac{64}{4}=16kg$

Report an Error

Example Question #61 : Fundamentals Of Force And Newton's Laws

A 10kg block accelerates at $7\frac{m}{s^2}$ across a sheet of ice. How much force was applied to the block?

Possible Answers:

F=70N

F=55N

F=90N

F=35N

Correct answer:

F=70N

Explanation:

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

Force pushing on a block

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.

Since the block is on ice, there is no friction.

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

Newton's second law is $F_{net}=ma$ .

Where $F_{net}$ is the net force applied in the direction of acceleration, is the mass of the block and is the acceleration.

The only force in the direction of acceleration is the applied force , therefore $F_{net}=F$ . The problem tells us m=10kg and $a=7\frac{m}{s^2}$ . Substituting in this information gives us:

F=ma=10*7=70N

Report an Error

Example Question #286 : Forces

A 8kg block accelerates at $3\frac{m}{s^2}$ across a sheet of ice. How much force was applied to the block?

Possible Answers:

F=70N

F=14N

F=24N

F=32N

Correct answer:

F=24N

Explanation:

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

Force pushing on a block

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.

Since the block is on ice, there is no friction.

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

Newton's second law is $F_{net}=ma$ .

Where $F_{net}$ is the net force applied in the direction of acceleration, is the mass of the block and is the acceleration.

The only force in the direction of acceleration is the applied force , therefore $F_{net}=F$ . The problem tells us m=8kg and $a=3\frac{m}{s^2}$ . Substituting in this information gives us

F=ma=8*3=24N

Report an Error

Example Question #61 : Fundamentals Of Force And Newton's Laws

A 132N force is applied to a 10kg block sitting on concrete. The coefficient of friction of concrete is 0.5 . What is the acceleration of the block?

Possible Answers:

$a=4.7\frac{m}{s^2}$

$a=9.2\frac{m}{s^2}$

$a=8.2\frac{m}{s^2}$

$a=2.5\frac{m}{s^2}$

Correct answer:

$a=8.2\frac{m}{s^2}$

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 ,

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.

Solving for acceleration, we have

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

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,

$F_{net}=F-f=132N-50N=82N$ . The problem tells us m=10kg . Substituting in this information gives us

$a=\frac{F_{net}}{m}=\frac{82}{10}=8.2\frac{m}{s^2}$

Report an Error

Example Question #285 : Forces

A 90N force is applied to a 30kg block sitting on wood. The coefficient of friction of wood is 0.2 . What is the acceleration of the block?

Possible Answers:

$a=2\frac{m}{s^2}$

$a=5\frac{m}{s^2}$

$a=1\frac{m}{s^2}$

$a=8\frac{m}{s^2}$

Correct answer:

$a=1\frac{m}{s^2}$

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=30kg . There is no acceleration in the y-direction, so a=0 .

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

$F_{net}=ma\Rightarrow F_N-W=(30)*0\Rightarrow F_N-mg=(30)*0$

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

Assuming g=10

F_N=mg=(30)(10)=300N

Now that we have F_N , we can find the force due to friction. Given that $\mu=0.2$ and F_N=300N ,

$f=\mu *F_N=0.2*300N=60N$

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

Solving for acceleration, we have

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

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,

$F_{net}=F-f=90N-60N=30N$ . The problem tells us m=30kg . Substituting in this information gives us

$a=\frac{F_{net}}{m}=\frac{30}{30}=1\frac{m}{s^2}$

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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

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

Report an Error

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

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

Report an Error

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AP Physics 1 : Fundamentals of Force and Newton's Laws

Study concepts, example questions & explanations for AP Physics 1

All AP Physics 1 Resources

Example Questions

Example Question #281 : Forces

Example Question #281 : Forces

Example Question #41 : Newton's Second Law

Example Question #284 : Forces

Example Question #61 : Fundamentals Of Force And Newton's Laws

Example Question #286 : Forces

Example Question #61 : Fundamentals Of Force And Newton's Laws

Example Question #285 : Forces

Example Question #51 : Newton's Second Law

Example Question #51 : Newton's Second Law

All AP Physics 1 Resources

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