Newton's Second Law - AP Physics 1

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Question

During time period , a rocket ship deep in space of mass travels from to . During time period , the rocket fires. During time period , the rocket travels from to .

Time periods , , and all took

Determine the average force during time period .

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Answer

Finding initial momentum:

$\overrightarrow{P}=m\overrightarrow{v}$

$\overrightarrow{v}=\frac{<x_f-x_i,y_f-y_i>}{t}$

Combining equations:

$\overrightarrow{P}=m\frac{<x_f-x_i,y_f-y_i>}{t}$

Plugging in values:

$\overrightarrow{P}=2.5510^5\frac{<5000-0,5000-10000>}{10}$

$\overrightarrow{P_i}=<1.27510^8,-1.27510^8>\frac{kgm}{s}$

$\overrightarrow{P}=m\overrightarrow{v}$

$\overrightarrow{v}=\frac{<x_f-x_i,y_f-y_i>}{t}$

Combining equations:

$\overrightarrow{P}=m\frac{<x_f-x_i,y_f-y_i>}{t}$

Converting to and plugging in values:

$\overrightarrow{P}=2.5510^5\frac{<-30000+15000,-5000-0>}{10}$

$\overrightarrow{P}_f=<-3.82510^8,-1.27510^8>\frac{kgm}{s}$

Using

$Impulse=\Delta\overrightarrow{P}=\overrightarrow{P_f}-\overrightarrow{P_i}$ Plugging in values:

$\overrightarrow{\Delta P} =<-3.82510^8,-1.27510^8>-<1.27510^8,-1.27510^8>$

$\Delta\overrightarrow{P}=<5.110^8,0>\frac{kgm}{s}$

Using:

$\Delta \overrightarrow{P}=\overrightarrow{F}\Delta t$

Solving for $\overrightarrow{F}$ and plugging in values

$\overrightarrow{F}=\frac{<5.110^8,0>\frac{kgm}{s}}{10s}$

$\overrightarrow{F}=5.110^7\ \frac{\textup{m}}{\textup{s}}$

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Question

A 2000kg car with a velocity of collides head on with a 6000kg truck with a velocity of . Which vehicle experiences the greater force? Which experiences the greater acceleration?

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Answer

The car and the truck experience equal and opposite forces, but since the car has a smaller mass it will experience greater acceleration than the truck according to the equation F = ma.

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

A greater mass will decrease the acceleration.

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Question

A man of mass 50kg on the top floor of a skyscraper steps into an elevator. What is the man's weight as the elevator accelerates downward at a rate of $1.5 \frac{m}{s^2}$ ?

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

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Answer

Use Newton's second law to solve this problem.

When the elevator is not moving, we get

$F = mg = (50kg)\cdot (10\frac{m}{s^2}) = 500 N$

However, when the elevator is accelerating downward, the man appears to be lighter since the elevator is negating some of the force from gravity. Written as an equation, we have:

Where is acceleration due to gravity and is the acceleration of the elevator.

Putting in our values, we get:

$F = (50kg)\cdot (10\frac{m}{s^2} - 1.5\frac{m}{s^2}) = 425N$

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Question

A skydiver of mass 70kg has jumped out of a plane two miles above the surface of the earth. After 20 seconds, he has reached terminal velocity, meaning he is no longer accelerating. What is the force of the air on the skydiver's body?

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

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Answer

This question is testing your understanding of terminal velocity and Newton's second law. Since the skydiver is at terminal velocity, the force of the air is equal to the force of gravity, resulting in zero net force and thus no acceleration. We just need to calculate the force of gravity on the skydiver to find the force of the air:

$F_g=F_{air} = ma = (70kg)(10\frac{m}{s^2}) = 700 N$

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Question

A skydiver of mass is mid jump and has an instantaneous acceleration of $4\frac{m}{s^2}$ . What is the force exerted on the diver from the air?

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

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Answer

There are two forces in play in this scenario. The first is gravity, and the second is air resistance. Since they are opposing each other, we can write:

$F_{net} = F_g - F_{air}$

Substituting in Newton's second law, we get:

$ma_{inst} = mg - F_{air}$

Rearranging for the force of air resistance, we get:

$F_{air} = mg - ma_{inst}$

Plugging in our values from the problem statement:

$F_{air} = (50kg)(10\frac{m}{s^2})- (50kg)(4\frac{m}{s^2}) = 300 N$

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Question

A diver of 50kg jumps from a platform 20m high into a pool. If the diver decelerates at a constant rate to zero velocity in 0.8 seconds after hitting the water, what is the force that the water exerts on the diver?

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

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Answer

We can use the equation for conservation of energy to calculate the velocity of the diver as he hits the water:

Cancel out initial kinetic and final potential energies, and plug in our expressions:

$mgh_i = \frac{1}{2}mv_f^2$

Cancel out mass and rearranging for final velocity:

$v_f = \sqrt{2gh_i}$

Plug in our values:

$v_f = \sqrt{2(10\frac{m}{s^2})(20m)} = 20 \frac{m}{s}$

We know that the diver then decelerates from this velocity to zero in 0.8 seconds, so we can calculate the acceleration:

$a = \frac{v_f - v_i}{t} = \frac{20\frac{m}{s}}{0.8s} = 25\frac{m}{s^2}$

Then use Newton's second law to calculate the force on the diver:

$F = ma = (50kg)(25 \frac{m}{s^2}) = 1250N$

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Question

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 $2\frac{m}{s^2}$ ?

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

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Answer

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:

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

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Question

A constant force of 30N acts on a a 10kg box as shown in the diagram. If the box is originally at rest, what will be its velocity after 5s?

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Answer

The box has a constant force acting on it pulling it towards the left. Therefore we can write this as:

, where the negative sign indicates that the force is directed towards the left.

Since the force is constant, this means that is is causing the box to move with a constant acceleration that we can calculate using Newton's second law of motion:

$a=\frac{-30N}{10kg}= -3\frac{m}{s^2}$

Now that we know the acceleration, we can calculate the final velocity after 5 seconds:

$a=\frac{V_{f}-V_{i}}{t}$

Where and $V_{i}=0$ since the box is originally at rest.

So we have that

$-3\frac{m}{s^2}=\frac{V_{f}-0\frac{m}{s}}{5s}$

$V_{f}=(-3\frac{m}{s})*(5s)=-15 \frac{m}{s}$ .

Note that the negative sign indicates that the box is moving to the left.

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Question

If we apply a constant force of on a object that is located on a frictionless surface, what is the acceleration of the object?

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Answer

Use Newton's second law.

$a=\frac{100N}{10kg}$

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

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Question

A 10kg box is being pushed across a frictionless field by two people. The box is moving with an acceleration of $1\frac{m}{s^2}$ . What is the force applied by the weaker person if the stronger person can push twice as hard?

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Answer

The force applied by the weaker person can be calculated using Newton's second law, which states:

$\sum F=ma$

The net force is equal to the product of the mass of the object and the acceleration of the object. We were given the mass and acceleration of the object, but only the ratio of the applied forces:

Solve for , the applied force from the weaker person:

$x=\frac{10}{3}N$

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Question

Is it possible to have a non-zero number of forces acting on an object (of non-zero mass), yet the object doesn't acclerate?

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Answer

Newton's second law states that the net force, or the vector sum of all the forces acting on an object, equals the mass times the acceleration. So, it is possible to have forces act on an object without acceleration if the forces are oriented such that they vector sum to zero. An example would be a person sitting in a chair. Gravity and the normal force both act on the person. However, these forces are equal in magnitude and opposite in direction. So the person doesn't accelerate.

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Question

In which of the following situations can we claim the net force is zero on the object described?

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Answer

Acceleration requires a change in one of two quantities: speed and direction

The elevator is traveling only upwards and at a non changing speed. The accleration is zero and thus the net force must be zero by newton's second law.

The bird taking flight is accelerating from rest to some non-zero speed. Net force is non zero because the speed changes.

The child on the swing set changes direction due to traveling in a semi-circle. Also their speed will change as well. Net force is non zero because both the speed and direction change.

The car breaking to a stop is changing speed. Net force is non zero because the speed changes.

The airplane, while maintaining a constant speed, changes direction and therefore accelerates. Net force is non zero because the direction changes.

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Question

A proton has a mass that's about times the mass of an electron. Given that electrostatic forces between a proton and electron cause them to attract one another with the same force, what can you say about the acceleration of the two particles?

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Answer

For this, we have to know that:

, where is a force, is an object's mass, and is its acceleration.

The text tells us that the electrostatic force between the proton and electron will exert the same force, but that the mass of the proton is times the mass of the electron.

$F_{proton}=F_{electron}$ , where $F_{proton}$ is the force on the proton and $F_{electron}$ is the force on the electron.

To determine the effect on their acceleration:

$m_{proton}\cdot a_{proton}=m_{electron}\cdot a_{electron}$ , where $m_{proton}$ and $m_{electron}$ are the masses of the proton and electron respectively, and $a_{proton}$ and $a_{electron}$ are the acceleration of the proton and electron respectively.

Since we know that $m_{proton}=2000m_{electron}$

$2000m_{electron}\cdot a_{proton}= m_{electron}\cdot a_{electron}$

Solving for the acceleration of the proton:

$a_{proton}=\frac{a_{electron}}{2000}$

The acceleration of the proton is $\frac{1}{2000}$ of the acceleration of the electron.

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Question

If a object is subjected to a force of , by how much will it accelerate?

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Answer

In this question, we're being told that an object of a given mass is being subjected to a force. To solve this problem, we'll need to make use of Newton's second law, which states that an object of a given mass will accelerate at a rate that is proportional to the force that is applied. Or, written in equation form:

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

Plugging in the values given, we obtain:

$a=\frac{10: N}{5: kg}=2: \frac{m}{s^{2}}$

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Question

2 objects(named object A and object B) of equal masses and initial kinetic energy collide onto one another. During the collision, object A loses $\frac{1}{2}$ of its kinetic energy, which object B gains. Assume mass of both objects remain unchanged.

Given that object A's mass is and its velocity changes by $3 \frac{m}{s}$ over a period of seconds, determine the average force applied on object A.

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Answer

Force is given by:

$F_{net}=ma$ , where is mass and is acceleration.

We're given mass, but we aren't given acceleration. Since the question asks for average force $F_{average}$ , we can determine average acceleration $a_{aver}$ .

$a_{aver}=\frac{\Delta v}{\Delta t}$ , where $\Delta v$ is the change in velocity and $\Delta t$ is change in time.

In our problem, $\Delta v= 3\frac{m}{s}$ and $\Delta t= 3s$

$a_{aver}=\frac{\Delta v}{\Delta t}=\frac{3\frac{m}{s}}{3s}=1\frac{m}{s^2}$

$F_{average}=ma_{average}=5kg*1\frac{m}{s^2}=5N$

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Question

A block with a mass of is pushed across a frictionless surface with a force of for a time of . What is the velocity of the block after the push?

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Answer

Here we must use the following formula:

We can substitute our known values of mass and force and solve for acceleration

$a = \frac{F}{m}= \frac{8N}{4kg}= 2\frac{m}{s^2}$

Since we know the acceleration and the time it acts upon the object, we can determine the final velocity through the following equation:

$v=at=(2\frac{m}{s^2})(3s)= 6\frac{m}{s}$

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Question

Consider a block sitting at rest on an inclined plane. Find the maximum inclination angle the plane may have without the block sliding if the coefficients of kinetic and static friction are $\mu_k=0.2,\mu_s=0.4$ , respectively.

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Answer

If the block is to remain at rest on the plane, we know that the sum of the forces acting a long the plane must be equal and opposite. This means that the gravitational force acting along the plane is equal to and opposite of the force of friction. This can be demonstrated as:

$mg\sin{\theta}=\mu_sN=\mu_smg\cos{\theta}$

This can be rewritten as:

$\tan{\theta}=\mu_s$

For the block to remain at rest, the force of static friction must exceed (for this problem we will set them equal to each other since it gives us the best approximation); solve for the angle:

$\tan{\theta}=\mu_s$

$\theta=\tan{^{-1}\mu_s}$

$\theta=\tan{^{-1}\(0.4)}=21.8^{\circ}$

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Question

A force of is applied to a object in space.

What is the acceleration of the object?

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Answer

Newton's second law states:

Where is the net force exerted upon an object, is the mass of the object and is the acceleration of the object.

We rearrange this equation to show:

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

Plug in our given values with and :

$\frac{15N}{5kg}=3\frac{m}{s^2}$

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Question

If of force is continuously applied to a box with mass , what will the box's velocity be after given that it's initial velocity was ?

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Answer

By Newton's second law:

, where is force, is mass, and is acceleration.

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

This is the acceleration. Since we're assuming this acceleration is constant over time, we can model velocity as:

where is the initial velocity.

Since the initial velocity is in our problem,

After seconds,

$v=2\frac{m}{s^2}*5s=10\frac{m}{s}$

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Question

A train of mass $5.11*10^{11}kg$ goes from $50 \frac{km}{hr}$ to $0 \frac{km}{hr}$ in . Calculate the deceleration in terms of .

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Answer

Use work:

All energy will be kinetic.

Convert $\frac{km}{hr}$ to $\frac{m}{s}$ :

$\frac{50km}{hr}\frac{1000m}{1km}\frac{1 hr}{3600 s}=13.9\frac{km}{hr}$

$\frac{0km}{hr}\frac{1000m}{1km}\frac{1 hr}{3600 s}=0\frac{km}{hr}$

Plug in values. Force will be negative as it is pointing against the direction of travel:

$F1000=.55.1110^{11}0^2-.55.1110^{11}(13.9)^2$

Solve for :

$F=-4.9310^{10}$

$|F|=4.9310^{10}$

Use Newton's second law:

Plug in values:

$5.1110^{11}a=4.9310^{10}$

Solve for :

$a=\frac{.096m}{s^2}$

Convert to

$\frac{.096m}{s^2}*\frac{g}{\frac{9.8m}{s^2}}=.0098g$

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