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AP Physics 1 : Newton's Second Law

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Example Questions

Example Question #44 : Fundamentals Of Force And Newton's Laws

A $205\textup{ lb}$ baseball player slides, making contact with the ground causing a force of kinetic friction measured at $250\textup{ N}$ . What is the acceleration that acts on this superstar due to friction during the slide?

$1\textup{ lb}=0.453\textup{ kg}$

Possible Answers:

$-3.6\ \frac{\textup{m}}{\textup{s}^{2}}$

$-2.7\ \frac{\textup{m}}{\textup{s}^{2}}$

$-3.0\ \frac{\textup{m}}{\textup{s}^{2}}$

$-3.3\ \frac{\textup{m}}{\textup{s}^{2}}$

Correct answer:

$-2.7\ \frac{\textup{m}}{\textup{s}^{2}}$

Explanation:

Firstly, put the given values into the proper units.

$205lbs=\frac{0.453Kg}{1lbs}=92.87Kg$

Now use Newton's second law to determine the acceleration.

F=ma

250N=92.87Kg(a)

$a=2.69\frac{m}{s^{2}}$

He will decelerate at $2.7\frac{m}{s^{2}}$ .

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Example Question #41 : Fundamentals Of Force And Newton's Laws

Kaden lives on the 4th floor at an apartment complex. Instead of making multiple trips up and down the stairs he elects to use a rope and bucket system to get his groceries up to his balcony. If Kaden applies $98\textup{ N}$ of tension on the rope tied to the bucket. What mass (in kilograms) of groceries can he carry in each load? Assume that both the bucket and rope are weightless and neglect friction.

Possible Answers:

$100\textup{ kg}$

$5\textup{ kg}$

$10\textup{ kg}$

$20\textup{ kg}$

Correct answer:

$10\textup{ kg}$

Explanation:

The bucket is not accelerating in either direction so the tension force must be equal to the weight of the groceries.

T=W

Multiplying mass and gravity gives weight

T=mg

Plug in and solve for the mass

$98N=m(9.8\frac{m}{s^{2}})$

m=10Kg

Kaden can carry a maximum of 10Kg of groceries in each load up to his balcony.

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Example Question #51 : Fundamentals Of Force And Newton's Laws

A pitcher throws a fastball at $100\ \frac{\textup{mi}}{hr}$ and the catcher brings the ball to a complete rest over a distance of $0.150\textup{ m}$ . What force did the catcher exert on the baseball to bring it to rest? $\textup{m}_{\textup{baseball}}=0.145\textup{ kg}$ .

$1\textup{ in}=2.54\textup{ cm}$

Possible Answers:

$1730\textup{ N}$

$1930\textup{ N}$

$1830\textup{ N}$

$1630\textup{ N}$

Correct answer:

$1930\textup{ N}$

Explanation:

First, convert to metric units.

$\frac{100mi}{hr}=\frac{5280ft}{1mi}=\frac{12in}{1ft}=\frac{2.54cm}{1in}=\frac{1m}{100cm}=\frac{1hr}{60min}=\frac{1min}{60s}=44.7\frac{m}{s}$

Now, use the equation for velocity.

$v=\frac{\Delta x}{\Delta t}$

Plug in and solve for the final time.

$44.7\frac{m}{s}=\frac{0.150m-0m}{t_{f}-0s}$

$t_{f}=0.0034s$

Now use the equation for acceleration.

$a=\frac{\Delta v}{\Delta t}$

Plug in and solve for the acceleration.

$a=\frac{0\frac{m}{s}-44.7\frac{m}{s}}{0.0034s-0s}$

$a=13320.6\frac{m}{s^{2}}$

Use Newton's second law to solve for the force.

F=ma

Plug in and solve for the force that the catcher exerts on the ball.

$F=0.145Kg(13320.6\frac{m}{s^{2}})$

F=1930N

The catcher will exert an amazing 1930N of force to stop the fastball in such a short distance.

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Example Question #32 : Newton's Second Law

Cody pulls his $65\textup{ kg}$ brother on the fresh snow during Christmas break. In order for his brother to have fun he must pull him at a constant speed of $6.0\ \frac{\textup{m}}{\textup{s}}$ . The coefficient of kinetic friction between the sled bottom and the fresh powder is 0.60 . What horizontal pulling force must Cody apply to give his brother a fun ride? Assume the sled has no mass and neglect air resistance.

Possible Answers:

$360\textup{ N}$

$380\textup{ N}$

$340\textup{ N}$

$320\textup{ N}$

Correct answer:

$380\textup{ N}$

Explanation:

The weight is being compensated by the normal force.

$F_{n}=w$

Weight is equal to mass times the acceleration due to gravity.

$F_{n}=mg$

$F_{n}=65Kg(9.8\frac{m}{s^{2}})$

$F_{n}=637N$

The sled does not have to accelerate so the kinetic frictional force is equal to the pulling force .

$P=F_{k}$

The equation for kinetic frictional force is inserted below.

$P=\mu _{k}F_{n}$

Plug in and solve for the pulling force.

P=0.60(637N)

P=380N

Cody must apply at least 380N of horizontal pulling force.

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Example Question #31 : Newton's Second Law

In the military a paratrooper and his/her gear can weigh in excess of $1100\textup{ N}$ . While skydiving, a paratrooper can experience an upward acceleration of $9.75\ \frac{\textup{m}}{\textup{s}^{2}}$ when he/she opens the parachute. What force does the paratrooper experience at this time?

Possible Answers:

$5050\textup{ N}$

$2190\textup{ N}$

$1090\textup{ N}$

$550\textup{ N}$

Correct answer:

$2190\textup{ N}$

Explanation:

The weight of the paratrooper and his or her gear is 1100N

w=mg

Plug in and solve for the mass.

$1100N=m(9.8\frac{m}{s^{2}})$

m=112.2Kg

The force due to air resistance causes an acceleration of $9.75\frac{m}{s^{2}}$ .

$F_{air}=ma$

Plug in and solve for the force due to air resistance.

$F_{air}=112.2Kg(9.75\frac{m}{s^{2}})$

$F_{air}=1090N$

The paratrooper is accelerating upward so we know that the force due to air resistance is greater than the force of gravity.

$F_{net}=w+F_{air}$

Plug in the known values and solve for the combined net force.

$F_{net}=1100N+1090N$ $F_{net}=2190N$

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Example Question #34 : Newton's Second Law

A new start up shoe company claims to make the lightest basketball shoe that still produces tremendous grip to allow players to move smoothly and freely throughout the game. In a laboratory test done on typical hard wood floor revealed that the $0.27\textup{ kg}$ shoe when swung at $15\ \frac{\textup{m}}{\textup{s}}$ takes $14\textup{ m}$ to come to a complete stop. Neglecting air resistance, what is the coefficient of kinetic friction between the shoe bottoms and the hard wood floor?

Possible Answers:

0.75

0.65

0.55

0.45

Correct answer:

0.45

Explanation:

Use the equation for velocity.

$v=\frac{\Delta v}{\Delta t}$

Plug in and solve for the final time.

$15\frac{m}{s}=\frac{14m-0m}{t_{f}-0s}$

$t_{f}=0.93s$

Now use the equation for acceleration.

$a=\frac{\Delta v}{\Delta t}$

Plug in and solve for the acceleration.

$a=\frac{15\frac{m}{s}-0\frac{m}{s}}{0.93s-0s}$

$a=-16.1\frac{m}{s^{2}}$

Now use Newton's second law.

$F_{k}=ma$

Plug in and solve for the kinetic frictional force.

$F_{k}=0.27Kg(-16.1\frac{m}{s^{2}})$

$F_{k}=4.34N$

Now solve for the normal force.

$F_{n}=w$

$F_{n}=mg$

Plug in and solve for the normal force.

$F_{n}=0.27Kg(9.8\frac{m}{s^{2}})$

$F_{n}=9.65N$

Now solve for the coefficient of kinetic friction.

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

Plug in and solve for the coefficient.

$4.34N=\mu _{k}(9.65N)$

$\mu _{k}=0.45$

The coefficient of kinetic friction between the floor and the bottom of these shoes was determined to be 0.45 .

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Example Question #31 : Newton's Second Law

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 0.35 . 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?

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

The sled and wood has a mass of 375Kg

w=mg

Plug ina and solve.

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

w=3675N

$w=F_{n}$

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.

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Example Question #31 : Newton's Second Law

During time period , a rocket ship deep in space of mass 2.55*10^5kg travels from <0,10>km to <5,5>km . During time period , the rocket fires. During time period , the rocket travels from <-15,0>km to <-30,-5>km .

Time periods , , and all took 10seconds.

Determine the average force during time period .

Possible Answers:

$2.3*10^7\ \frac{\textup{m}}{\textup{s}}$

$7.7*10^7\ \frac{\textup{m}}{\textup{s}}$

None of these

$5.1*10^7\ \frac{\textup{m}}{\textup{s}}$

$1.2*10^7\ \frac{\textup{m}}{\textup{s}}$

Correct answer:

$5.1*10^7\ \frac{\textup{m}}{\textup{s}}$

Explanation:

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:

$\overrightarrow{P}=2.55*10^5*\frac{<5000-0,5000-10000>}{10}$

$\overrightarrow{P_i}=<1.275*10^8,-1.275*10^8>\frac{kg*m}{s}$

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

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

Combining equations:

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

Converting to and plugging in values:

$\overrightarrow{P}=2.55*10^5*\frac{<-30000+15000,-5000-0>}{10}$

$\overrightarrow{P}_f=<-3.825*10^8,-1.275*10^8>\frac{kg*m}{s}$

Using

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

$\overrightarrow{\Delta P} =<-3.825*10^8,-1.275*10^8>-<1.275*10^8,-1.275*10^8>$

$\Delta\overrightarrow{P}=<5.1*10^8,0>\frac{kg*m}{s}$

Using:

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

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

$\overrightarrow{F}=\frac{<5.1*10^8,0>\frac{kg*m}{s}}{10s}$

$\overrightarrow{F}=5.1*10^7\ \frac{\textup{m}}{\textup{s}}$

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Example Question #31 : Newton's Second Law

Which of the following scenarios is an example of no net force being involved?

Possible Answers:

A mass attached to a pendulum moves back and forth with constant frequency

A box is lowered at a constant speed

A car gradually slows down as it approaches a red light

An apple falls from a tree

A person on a ferris wheel moves at a constant speed

Correct answer:

A box is lowered at a constant speed

Explanation:

In this question, we're being asked to identify a situation in which no net force is involved.

In order for there to be no net force on an object, then that object must not be accelerating. Hence, we need to look at each answer choice to see if there is any acceleration involved.

For a car that is gradually slowing down, a net force will be involved. This is because the car is undergoing a negative acceleration; it is accelerating in the direction directly opposite to its direction of motion.

In the case of an apple falling from a tree, a net force is also involved. This is because as the apple falls, it is being subjected to the acceleration due to gravity.

A mass attached to a pendulum will also experience a net force. This is because the pendulum is constantly going back and forth, meaning that the direction of acceleration is constantly going back and forth.

Since the motion of a person riding a ferris wheel is circular, there is a continuously changing velocity. Even if speed is kept constant, the direction of motion is always changing, thus the rider experiences a net acceleration and thus a net force.

Finally, a box being lowered at a constant rate will not experience a net force. This is because the box has a constant velocity, which means that no acceleration is involved.

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Example Question #847 : Newtonian Mechanics

A quarterback throws a football a horizontal distance of 50 yd to a wide receiver. The ball was airborne for 5 s . The ball had an initial speed of 50 mph . The ball has a mass of .43kg .

$1mph=.447\frac{m}{s}$

1 yd=.914 m

If the quarterback’s throw took .75 s , determine the average force of the throw.

Possible Answers:

<5.5,11.5>N

<9.8,9.8>N

<2.6,3.7>N

<9.5,5.5>N

None of these

Correct answer:

<5.5,11.5>N

Explanation:

Determining horizontal component of velocity:

d_x=v_x*t

$v_x=\frac{d_x}{t}$

$v_x=\frac{45.7}{5}$

$v_x=9.14\frac{m}{s}$

Using

v_x^2+v_y^2=v^2

Solving for v_y

$v_y=\sqrt{v^2-v_x^2}$

Combining equations

$v_y=\sqrt{v^2-(\frac{d_x}{t})^2}$

Converting mph to mps

$50mph*\frac{.447mps}{1 mph}=22.35\frac{m}{s}$

Converting yards to meters

$50yards*\frac{.914 meters}{1 yard}=45.7 meters$

Plugging in values:

$v_y=\sqrt{v^2-(\frac{d_x}{t})^2}$

$v_y=\sqrt{22.35^2-(\frac{45.7}{5})^2}$

$v_y=20.4\frac{m}{s}$

$\overrightarrow{v}=<9.14,20.4>\frac{m}{s}$

Using

v_f=v_i+a*t

$<9.14,20.4>\frac{m}{s}=<0,0>+<a_x,a_y>*.75$

$<a_x,a_y>=<12.8,26.7>\frac{m}{s}$

Using

F=ma

$F=.43kg*<12.8,26.7>\frac{m}{s}$

F=<5.5,11.5>N

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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 #44 : Fundamentals Of Force And Newton's Laws

Example Question #41 : Fundamentals Of Force And Newton's Laws

Example Question #51 : Fundamentals Of Force And Newton's Laws

Example Question #32 : Newton's Second Law

Example Question #31 : Newton's Second Law

Example Question #34 : Newton's Second Law

Example Question #31 : Newton's Second Law

Example Question #31 : Newton's Second Law

Example Question #31 : Newton's Second Law

Example Question #847 : Newtonian Mechanics

All AP Physics 1 Resources

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