Physics: Newton's Laws and Mechanics

Practice flashcards reviewing Newton's First, Second, and Third Laws, kinematics, mass vs. weight, normal forces, and contact forces based on lecture ConcepTests.

According to Newton's First Law, why does a book lying at rest on a table remain at rest?

There is no net force acting on the book because the downward force of gravity and the upward force exerted by the table are equal in magnitude and cancel each other out.

What is the net force acting on a hockey puck sliding on ice at a constant velocity?

The net force is $0$ because the puck is moving at a constant velocity and therefore has zero acceleration.

Why does a book slide forward off a bus seat when the bus stops suddenly?

No net force acted on the book; it was initially moving forward with the bus and simply continued its initial state of motion forward.

When a smooth flat stone kicked onto a frozen pond slows down and eventually stops, what can be concluded about the net force acting on it?

A net force acted on it all along; friction acted on the stone to slow it down and stop it.

What happens to a cart on a horizontal frictionless table once it has been given an initial push and released?

The cart will continue moving with constant velocity because there is no longer a horizontal force acting on it after release.

What must be done to make a cart on a horizontal frictionless track continue moving with constant acceleration?

The cart must be pushed continuously to maintain a non-zero applied force as it moves down the track.

What happens to a very large truck on a frictionless frozen lake when a fly smashes against its front window?

The truck accelerates backward during the brief impact and then continues moving backward at constant speed once the force ceases.

If a force $F$ accelerates a Ferrari from rest to speed $v$ in $4\,\text{s}$, how long must a force of $\frac{1}{2}F$ be applied to reach the same final speed?

It must be applied for $8\,\text{s}$ because halving the force halves the acceleration, requiring double the time to reach final velocity ($v = aT$).

Why does an automobile's odometer not measure zero after a long trip that returns home?

The odometer records total distance traveled rather than displacement, which is dependent only on the net change in position.

Starting from rest, a force $F$ moves a Ferrari $50\,\text{m}$ in $4\,\text{s}$. How far would the car travel if the same force were applied for $8\,\text{s}$?

It would travel $200\,\text{m}$ because distance varies with the square of time ($x = \frac{1}{2}aT^2$), meaning doubling the time quadruples the distance.

A braking force $F$ applied for $4\,\text{s}$ slows a Ferrari over $25\,\text{m}$ without stopping it. How far will the car travel in total if the force is applied for $8\,\text{s}$?

The total distance traveled will be $25\,\text{m} < x < 50\,\text{m}$ because the car slows down and covers less distance during the second $4\,\text{s}$ than the initial $25\,\text{m}$.

A force $F$ acts on mass $M$ for time $T$ giving final speed $v$. What is the final speed if the same force acts for time $T$ on mass $2M$?

The final speed is $\frac{1}{2}v$ because doubling the mass cuts the acceleration in half, producing half the final speed in the same time.

A force $F$ gives mass $m_1$ acceleration $a_1$ and mass $m_2$ acceleration $a_2 = 2a_1$. What acceleration does $F$ produce if both masses are glued together?

The resulting acceleration is $\frac{2}{3}a_1$ because $m_2 = \frac{1}{2}m_1$, giving a combined mass of $\frac{3}{2}m_1$, so $a = \frac{F}{\frac{3}{2}m_1} = \frac{2}{3}a_1$.

If two people can each pull with a force of $20\,\text{N}$, what is the best way to rip a rope in half?

Tie one end of the rope to a tree and team up to pull together on the other end, allowing the tree to exert an equal reacting force of $40\,\text{N}$.

According to Newton's Third Law, how does pulling down on a rope allow a climber to ascend?

Pulling down on the rope causes the rope to exert an equal and opposite upward reaction force on the climber, propelling them upward.

In outer space, how do the gravitational forces exerted by a bowling ball and a Ping-Pong ball on each other compare?

The attractive gravitational forces are equal in magnitude and opposite in direction according to Newton's Third Law.

When a bowling ball and a Ping-Pong ball attract each other gravitationally in space, how do their accelerations compare?

The accelerations are opposite in direction, but the Ping-Pong ball experiences a greater acceleration because $a = \frac{F}{m}$ and it has a smaller mass.

In a collision between a small car and a large truck, which vehicle experiences the greater impact force?

Both vehicles experience the exact same magnitude of force according to Newton's Third Law.

In a collision between a small car and a large truck, which vehicle experiences the greater acceleration?

The car experiences the greater acceleration because both vehicles feel the same force magnitude, and acceleration is inversely proportional to mass ($a = \frac{F}{m}$).

Two blocks of masses $2m$ and $m$ are in contact on a frictionless surface. If a force $F$ is applied to mass $2m$, what is the contact force on mass $m$?

The contact force is $\frac{1}{3}F$ because the entire system accelerates at $a = \frac{F}{3m}$, requiring a force of $m \times a = \frac{1}{3}F$ on mass $m$.

How does the gravitational force $F_g$ acting on a stone compare to that acting on a feather?

$F_g$ is greater on the stone because gravitational force depends directly on mass ($F_g = mg$), and the stone has greater mass.

In a vacuum, how does the gravitational acceleration of a stone compare to that of a feather?

The acceleration is equal on both objects ($g$) because mass cancels out in the equation $a = \frac{F}{m} = \frac{mg}{m}$.

If an astronaut kicks a bowling ball on Earth and hurts their foot, how does their foot feel if they kick it with the same force on the Moon?

Their foot hurts the same because the bowling ball and the astronaut maintain the exact same mass and inertia on the Moon.

For identical blocks of weight $W$, one resting on a flat surface (Case A) and one on an incline of angle $\theta$ (Case B), which has the larger normal force?

The normal force is greater in Case A because on a flat surface $N = W$, whereas on an incline $N = W \cos(\theta) < W$.