In this problem, a force $F = 100~mathrm{N}$ acts horizontally on a chair of mass $m = 8.0~mathrm{kg}$. The chair moves with constant speed. We find which statement is true.

The speed of the object is constant, so the acceleration is $0$. The net force must be 0, so the magnitude of the frictional force must be equal to the applied force. The true statement is $textbf{C.}$ “the friction force is $100~mathrm{N}$.

There is nothing that pushes the puck, so the force exerted on it must be zero and there is no acceleration. Hence, the speed is constant. The correct statement is $textbf{B.}$ “the puck moves at constant speed.”

The only two forces along the vertical is the weight and the normal force, since the desktop is level so they must have the same magnitude but opposite directions. Hence, the answer is $textbf{A.}$ the weight of the book.

Consider the axes to the the ramp and a line perpendicular to the ramp. The net force is zero. Let downwards down the ramp and upwards be the positive directions.

$$
begin{align}
sum F_{x} &= 0 = W sin theta – T – F_text{f} \
sum F_{y} &= 0 = N – W cos theta
end{align}
$$

Rearranging equations (1) and (2) gives

$$
begin{align*}
W sin theta &= T + F_text{f} \
N &= W cos theta
end{align*}
$$

We see that $textbf{C.}$ is the correct relationship.

In this problem, we are given the same setup as problem 4. From the previous item, we have

$$
begin{align*}
W sin theta &= T + F_text{f} \
N &= W cos theta
end{align*}
$$

We find what happens to $N$, $F_text{f}$, and $T$ if $theta$ is increased.

We also know that
$$
F_text{f} = mu N = mu W cos theta
$$
The relationships can be rewritten as

$$
begin{align*}
N &= W cos theta tag{1} \
F_text{f} &= mu W cos theta tag{2} \
T &= W sin theta – mu W cos theta tag{3}
end{align*}
$$

For the normal force, in equation (1), as $theta$ increases, $cos theta$ would decrease, so the normal force $textbf{decreases}$.

For the frictional force, in equation (2), as $theta$ increases, $cos theta$ would decrease, so the frictional force $textbf{decreases}$.

For the tension, in equation (3), as $theta$ increases, $sin theta$ would increase and $-cos theta$ would also increase, so the tension $textbf{increases}$. The correct choice is $textbf{D}$. $N$ decreases, $F_text{f}$ decreases, and $T$ increases.

From the force-equation, we have

$$
begin{align*}
F &= ma
end{align*}
$$

The mass is the slope of the $F$ vs $a$ graph. The graph starts at the origin. At $a = 5~mathrm{m/s^{2}}$, $F = 3~mathrm{N}$. The slope is
$$
boxed{ mathrm{B.~} 0.6~mathrm{kg} }
$$

In this problem, we are given the force vs acceleration graph of a cart pulled across a level surface. When a force of $F = 4~mathrm{N}$ is applied, the car accelerates for $a = 5~mathrm{m/s^{2}}$. We find the frictional force.

At $a = 5~mathrm{m/s^{2}}$, the net force must be $F_text{net} = 3~mathrm{N}$. To find the frictional force $f$, we have

$$
begin{align*}
F_text{net} &= F – f \
implies f &= F – F_text{net} \
&= 4~mathrm{N} – 3~mathrm{N} \
f &= 1~mathrm{N}
end{align*}
$$

The frictional force is
$$
boxed{ mathrm{B.~} 1.0~mathrm{N} }
$$

In this problem, a block of mass $m = 1.6~mathrm{kg}$ is pulled across a level surface by a horizontal force of $F = 12~mathrm{N}$. We calculate the approximate coefficient of kinetic friction between the block and the surface if the block moves with constant velocity. We use $g = 9.81~mathrm{m/s^{2}}$

The velocity is constant, so the acceleration and net force must be zero. The frictional force is $f = mu_text{k} mg$. We have

$$
begin{align*}
F_text{net} &= F – f \
0 &= F – mu_text{k} mg \
implies mu_{k} &= frac{F}{mg} \
&= frac{12~mathrm{N}}{left( 1.6~mathrm{kg} right) left( 9.81~mathrm{m/s^{2}} right)} \
&= 0.76453 \
mu_{k} &= 0.76
end{align*}
$$

The approximate coefficient of friction is approximately
$$
boxed{ mathrm{C.~} 0.75~ }
$$

In this problem, when a box is kicked, it slides $Delta x_{1} = 3~mathrm{m}$ across the floor before coming to a stop. After a second kick, the distance covered is $Delta x_{2} = 6~mathrm{m}$. We compare the friction forces and initial velocities of the two cases.

The weight of the box does not change, so the normal force is constant. The friction force is proportional to the normal force. Therefore, the friction force of the box is $textbf{the same}$ for the two cases.

To find the initial velocities, we find the relationship of $v_text{i}$ and $Delta x$. When the box stops, its velocity is $v_text{f} = 0$. Notice that the acceleration is the same for both cases since the friction force is the same.

$$
begin{align*}
v_text{f}^{2} &= v_text{i}^{2} + 2a Delta x \
0 &= v_text{i}^{2} + 2a Delta x \
implies v_text{i}^{2} &= -2a Delta x \
v_text{i} &= sqrt{-2a} sqrt{Delta x} \
v_text{i} &propto sqrt{Delta x}
end{align*}
$$

The initial velocity is proportional to the square root of the stopping distance. The stopping distance of the second kick is twice of the first kick, so its initial speed in the second kick must be $sqrt{2}$ times the initial speed in the first kick.