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

Which of the following is not one of Newton's fundamental laws?

In order for an object to be in motion, a force needs to act on it

An object in motion will remain in that state of motion unless an outside force acts on that object

The acceleration experienced by an object is directly proportional to the force applied to that object, and inversely proportional to the mass of that object

For every action, there will always be an opposite reaction equal in magnitude

All of these correctly represent Newton's laws

Explanation

To answer this question, we'll need to be familiar with Isaac Newton's fundamental laws of physics, which forms the basis of classical mechanics.

Newton's first law states that an object in motion will remain in that state of motion indefinitely, unless an outside force acts on this object.

Newton's second law can be stated as follows:

This expression tells us that the acceleration experienced by an object is directly proportional to the applied force and inversely proportional to the object's mass.

Newton's third law states that for every action, there is always an equal and opposite reaction. Another way of saying this is that if one object were to exert a force on a second object, then the second object would exert the same force against the first object.

One of the answer choices states the following

"In order for an object to be in motion, a force needs to act on it."

While this may seem intuitive based on our everyday experience, and it may also show some resemblance to Newton's first law, this statement is actually incorrect. Force is not responsible for motion - force is responsible for changes in motion.

Which of the following is not one of Newton's fundamental laws?

In order for an object to be in motion, a force needs to act on it

An object in motion will remain in that state of motion unless an outside force acts on that object

The acceleration experienced by an object is directly proportional to the force applied to that object, and inversely proportional to the mass of that object

For every action, there will always be an opposite reaction equal in magnitude

All of these correctly represent Newton's laws

Explanation

To answer this question, we'll need to be familiar with Isaac Newton's fundamental laws of physics, which forms the basis of classical mechanics.

Newton's first law states that an object in motion will remain in that state of motion indefinitely, unless an outside force acts on this object.

Newton's second law can be stated as follows:

This expression tells us that the acceleration experienced by an object is directly proportional to the applied force and inversely proportional to the object's mass.

One of the answer choices states the following

"In order for an object to be in motion, a force needs to act on it."

Dave is riding his skateboard and pushes off the ground with his foot. This causes him to accelerate at a rate of $4\frac{m}{s^2}$ . Dave weighs 589 N. How strong was his push off the ground?

Explanation

Dave weighs 589 N. This means his mass is

$\frac{589N}{9.81\frac{m}{s^2}}=60.0kg$

He accelerates at $4\frac{m}{s^2}$ , which means he was pushed by a force of

$F=ma=(60kg)(4.0 \frac{m}{s^2})=240N$

By Newton's third law of motion, Dave must also have pushed of the ground with a force of 240 Newtons.

Dave is riding his skateboard and pushes off the ground with his foot. This causes him to accelerate at a rate of $4\frac{m}{s^2}$ . Dave weighs 589 N. How strong was his push off the ground?

Explanation

Dave weighs 589 N. This means his mass is

$\frac{589N}{9.81\frac{m}{s^2}}=60.0kg$

He accelerates at $4\frac{m}{s^2}$ , which means he was pushed by a force of

$F=ma=(60kg)(4.0 \frac{m}{s^2})=240N$

By Newton's third law of motion, Dave must also have pushed of the ground with a force of 240 Newtons.

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

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 where is the mass of the block and is the gravitational constant.

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 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, and . There are in opposite directions, so they are subtracted. We are given . There is no acceleration in the y-direction, so .

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

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

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

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 and $a=3\frac{m}{s^2}$ . Substituting this information into Newton's second law gives us

$F_{net}=ma$

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

A 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?

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 where is the mass of the block and is the gravitational constant.

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

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

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

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

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

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

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

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 where is the mass of the block and is the gravitational constant.

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 . Substituting in this information gives us

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

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

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 where is the mass of the block and is the gravitational constant.

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

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

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

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

John has just finished chopping a bunch of firewood and has to get it in a pile by his house. He elects to use a sled to pull the wood across the gravel road to his house. The road and the sled maintain a coefficient of static friction determined to be . John is able to cram $375\textup{ kg}$ of wood into the sled. After chopping all the wood he is tired but convinced he can move the sled. If John can pull with a max force of $645\textup{ N}$ can he pull the sled to his house and if not how many newtons of force is he short?