"Weightlessness" is analogous to free-fall (neglecting air resistance), during which the only force on an object is the force of gravity. When normal force becomes zero, the object loses physical contact with the surface.

When normal force is equal to the force of gravity, the object is in equilibrium and the net vertical force is zero. This results in zero acceleration, but does not result in "weightlessness." Mass is not a force, and negative force values merely indicate relative direction (assuming correct calculations)

Frictional force is calculated using the normal force and the friction coefficient. Since the book is at rest (not in motion), we use the coefficient of static friction.

\(\displaystyle F_f=\mu_sF_N\)

The normal force will be equal and opposite the force pushing directly into the surface; in this case, that means the normal force will be 20N.

Solve for the force of friction:

\(\displaystyle F_f=(0.3)(20N)=6N\)

\(\displaystyle \begin{align*}&\text{To solve a problem of force balance, a good idea is to draw}\\&\text{it out. Since the block is at rest, i.e. stationary, the sum}\\&\text{of forces acting on it must be equal to zero. The forces acting}\\&\text{on it are gravity (in the form of the mass times the gravitational}\\&\text{constant), the force of the rope supporting it, a normal force,}\\&\text{and since it’s on an incline, a frictional force to balance}\\&\text{it all out. Notice that the tension in the rope mitigates some}\\&\text{of the gravitational force (as illustrated in the upper left}\\&\text{corner) to lessen what is applied to the slope:}\\&F_{mass}=m*g\\&F_{normal}=(F_{mass}-F_{rope})*cos(a)\\&F_{friction}=(F_{mass}-F_{rope})*sin(a)\\&F_{normal}=(190kg(9.81\frac{m}{s^2})-559.17N)(cos(33^{\circ}))\\&F_{normal}=1094.24N\end{align*}\)

\(\displaystyle \begin{align*}&\text{To solve a problem of force balance, a good idea is to draw}\\&\text{it out. Since the block is at rest, i.e. stationary, the sum}\\&\text{of forces acting on it must be equal to zero. The forces acting}\\&\text{on it are gravity (in the form of the mass times the gravitational}\\&\text{constant), a normal force, and since it’s on an incline, a frictional}\\&\text{force to balance it all out:}\\&F_{mass}=m*g\\&F_{normal}=m*g*cos(a)\\&F_{friction}=m*g*sin(a)\\&F_{normal}=233kg(9.81\frac{m}{s^2})(cos(81^{\circ}))\\&F_{normal}=357.57N\end{align*}\)

\(\displaystyle \begin{align*}&\text{To solve a problem of force balance, a good idea is to draw}\\&\text{it out. Since the block is at rest, i.e. stationary, the sum}\\&\text{of forces acting on it must be equal to zero. The forces acting}\\&\text{on it are gravity (in the form of the mass times the gravitational}\\&\text{constant), a normal force, and since it’s on an incline, a frictional}\\&\text{force to balance it all out:}\\&F_{mass}=m*g\\&F_{normal}=m*g*cos(a)\\&F_{friction}=m*g*sin(a)\\&F_{normal}=187kg(9.81\frac{m}{s^2})(cos(83^{\circ}))\\&F_{normal}=223.57N\end{align*}\)

\(\displaystyle \begin{align*}&\text{To solve a problem of force balance, a good idea is to draw}\\&\text{it out. Since the block is at rest, i.e. stationary, the sum}\\&\text{of forces acting on it must be equal to zero. The forces acting}\\&\text{on it are gravity (in the form of the mass times the gravitational}\\&\text{constant), a normal force, and since it’s on an incline, a frictional}\\&\text{force to balance it all out:}\\&F_{mass}=m*g\\&F_{normal}=m*g*cos(a)\\&F_{friction}=m*g*sin(a)\\&F_{normal}=267kg(9.81\frac{m}{s^2})(cos(49^{\circ}))\\&F_{normal}=1718.4N\end{align*}\)

\(\displaystyle \begin{align*}&\text{To solve a problem of force balance, a good idea is to draw}\\&\text{it out. Since the block is at rest, i.e. stationary, the sum}\\&\text{of forces acting on it must be equal to zero. The forces acting}\\&\text{on it are gravity (in the form of the mass times the gravitational}\\&\text{constant), the force of the rope supporting it, a normal force,}\\&\text{and since it’s on an incline, a frictional force to balance}\\&\text{it all out. Notice that the tension in the rope mitigates some}\\&\text{of the gravitational force (as illustrated in the upper left}\\&\text{corner) to lessen what is applied to the slope:}\\&F_{mass}=m*g\\&F_{normal}=(F_{mass}-F_{rope})*cos(a)\\&F_{rope}=F_{mass}-\frac{F_{normal}}{cos(a)}\\&F_{rope}=1098.72N-\frac{417.98N}{cos(18^{\circ})}\\&F_{rope}=659.23N\end{align*}\)

Relevant equations:

\(\displaystyle \Delta K = W\)

\(\displaystyle K = \frac{1}{2}mv^2\)

\(\displaystyle W = F*d\)

\(\displaystyle F_f = \mu F_N\)

Determine the truck's change in kinetic energy. Note that the final kinetic energy will be zero because it has no final velocity.

\(\displaystyle K_i = \frac{1}{2}(3500kg)(25\frac{m}{s})^2=1,093,750J\)

\(\displaystyle K_f=0\)

\(\displaystyle \Delta K = K_f - K_i = -1,093,750 J\)

Using the work-energy theorem, set change in kinetic energy equal to the work done by friction.

\(\displaystyle \Delta K = -F_f d\)

Substitute the equation for force of friction.

\(\displaystyle \Delta K=-\mu F_Nd\)

\(\displaystyle \Delta K = -\mu (mg)d\)

Use the given values for coefficient of friction, mass, and acceleration of gravity to solve for the distance.

\(\displaystyle -1,093,750 J= -(0.4)(3500kg)(9.8\frac{m}{s^2})d\)

\(\displaystyle d = 79.7m \approx 80m\)

The two horizontal forces acting on the box are the applied force of 40N to the right, and the friction force (\(\displaystyle \mu_kF_N\))to the left. So the net horizontal force is written as the equation below.

\(\displaystyle 40N-\mu_kF_N=ma\)

\(\displaystyle F_N=mg\) is the normal force. Solving for acceleration, a, we get the equation below.

\(\displaystyle a=\frac{40N-\mu_kF_N}{m}\)

We know the following from the question.

\(\displaystyle m=14kg\)

\(\displaystyle g=9.8\frac{m}{s^2}\)

\(\displaystyle \mu_k=0.2\)

\(\displaystyle F_N=mg=(14kg)(9.8\frac{m}{s^2})=137.2N\)

Thus, the acceleration is

\(\displaystyle a=\frac{40N-(0.2)(137.2N)}{14kg}\)

\(\displaystyle a=0.9\frac{m}{s^2}\)

Relevant equations: 

\(\displaystyle F_{net}= ma\)

\(\displaystyle F_f=\mu F_N\)

\(\displaystyle F_g = mg\)

Write net force equations for horizontal and vertical directions:

Horizontal: \(\displaystyle F_{net, x} = F_{pull, x}-F_f\)

Vertical: \(\displaystyle F_{net,y}=F_N+F_{pull,y}-F_g\)

Since the crate is not accelerating, \(\displaystyle \small a=0\frac{m}{s^2}\) in each direction, implying that \(\displaystyle \small F_{net,x}=F_{net,y}=0N\).

Horizontal: \(\displaystyle 0 = F_{pull,x}-F_{f}\)

Vertical: \(\displaystyle 0 = F_N + F_{pull,y}-F_g\)

Express the horizontal and vertical components of the pulling force in terms of the total pulling force, \(\displaystyle F_{pull}\).

Horizontal: \(\displaystyle 0 = (F_{pull}*cos(45^o))-F_f\)

Vertical: \(\displaystyle 0 = F_N + (F_{pull}*sin(45^o))-F_g\)

Replace force of friction with its expression in terms of the normal force.

Horizontal: \(\displaystyle 0 = (F_{pull}*cos(45^o))-(\mu F_N)\)

Rearrange to isolate the normal force.

Horizontal: \(\displaystyle F_N =\frac{F_{pull}*cos(45^o)}{\mu}\)

Substitute this term for the normal force in the net vertical force equation.

Vertical: \(\displaystyle 0 = (\frac{F_{pull}*cos(45^o)}{\mu})+(F_{pull}*sin(45^o))-F_g\)

Solve to isolate the pulling force.

\(\displaystyle F_{pull}=\frac{F_g*\mu}{cos(45^o)+\mu*sin(45^o)}\)

Use the given values for the mass of the box, coefficient of friction, and acceleration of gravity to solve for the pulling force.

\(\displaystyle F_{pull}=\frac{(15kg)(9.8 \frac{m}{s^2})(0.20)}{cos(45^o)+0.20*sin(45^o)}\)

\(\displaystyle F_{pull}=\frac{29.4N}{0.849}=34.65N\approx35N\)