AP Physics 1 - Newton's Second Law

A man is rolling a recycling bin of mass down his driveway, which has a slope of $5^{\circ}$ , at a constant velocity when he accidentally drops it. What is the total frictional force on the recycling bin if it is decelerating at a rate of $2\frac{m}{s^2}$ ?

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

Explanation

Since we are neglecting air resistance, there are two forces in play: gravity and friction. Therefore, we can use Newton's second law to write the following:

$F_{net} = ma$

Substituting in an expression for the force of gravity and rearranging for the force of friction, we get:

$F_f = mgsin(\theta)-ma = (5kg)(10\frac{m}{s^2})sin(5^{\circ})-(5kg)(-2\frac{m}{s^2})$

A block accelerates at $2\frac{m}{s^2}$ across concrete. The coefficient of friction of concrete 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

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

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

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

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

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

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

$F_{net}=ma$

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

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?

No, he would need $645\textup{ N}$ more.

Yes, the $645\textup{ N}$ will suffice to move the sled up to his house.

No, he would need $300\textup{ N}$ more.

No, he would need $500\textup{ N}$ more.

Explanation

The sled and wood has a mass of

Plug ina and solve.

$w=375Kg(9.8\frac{m}{s^{2}})$

$w=F_{n}$

Use the equation for static friction.

$F_{s}=\mu _{s}(F_{n})$

Plug in and solve for the static frictional force.

$F_{s}=0.35(3675N)$

$F_{s}=1286.25N$

John will not be able to move the sled and would have to double his efforts to move it.

A truck is pulling a trailer and is accelerating at $1\frac{m}{s^2}$ . The truck has mass and the trailer has mass . Determine the force and direction the trailer is applying on the truck.

, backwards

, forwards

, backwards

Explanation

The mass of the truck is irrelevant to this problem.

The trailer has a mass of , and is accelerating at $1\frac{m}{s^2}$ .

This is the force the truck is applying to the trailer to accelerate it forward. Based on Newton's Laws, the trailer is applying the same force, in the opposite direction.

Consider the following system:

Frictionless pully

Two masses, A and B, are attached to the end of a rope that runs through a pulley.

The system is at rest. The pulley has become quit old and has develop a thick coat of rust. If , what range of mass B will result in the system remaining at rest if the static frictional force within the pulley is ?

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

$2.5kg\leq m_B \leq 5.5kg$

$m_B\leq 4kg$

$m_B\geq 6kg$

$2kg\leq m_B \leq6kg$

$1kg\leq m_B \leq7kg$

Explanation

We will approach this problem by looking at the pulley. There are 3 forces applied to the pulley, tension from each weight as well as its internal friction force. To clarify, we will denote that forces in the counterclockwise direction will be positive and forces in the clockwise direction are negative. Also, we need to remember that the internal static friction can be applied in both the clockwise and counterclockwise direction. This will create 2 scenarios. Let's first began when friction is applied in the counter clockwise direction. We will start with Newton's 2nd law:

$F_{net}=ma$

Since we are asked to find when the system remains at rest, that means the net force on the system is 0:

$F_{net}=0$

Now we can begin looking at the three forces. First, we have tension created from mass A. This will be positive since it is in the counterclockwise direction relative to the pulley: