Newton's second law states that \(\displaystyle \vec{F}=m\vec{a}\).

In this problem, \(\displaystyle \vec{F}=1.5N\) and \(\displaystyle m=3.2kg\).

Plug these into the equation to solve for acceleration.

\(\displaystyle \vec{F}=m\vec{a}\)

\(\displaystyle 1.5N=3.2kg(a)\)

\(\displaystyle \frac{1.5N}{3.2kg}=a\)

\(\displaystyle 0.47\frac{m}{s^2}=a\)

Newton's second law states that \(\displaystyle \vec{F}=m\vec{a}\).

Plug in the values given to us and solve for the force.

\(\displaystyle \vec{F}=(1.12kg)(1.11\frac{m}{s^2})\)

\(\displaystyle \vec{F}=1.24N\)

Newton's second law states that \(\displaystyle \vec{F}=m\vec{a}\).

Plug in the given values to solve for the mass.

\(\displaystyle \vec{F}=m\vec{a}\)

\(\displaystyle 31N=m(2.1\frac{m}{s^2})\)

\(\displaystyle \frac{31N}{2.1\frac{m}{s^2}}=m\)

\(\displaystyle 14.76kg=m\)

Newton's second law states that \(\displaystyle \vec{F}=m\vec{a}\).

If Derek is pushing with \(\displaystyle 35N\) of force, then we should be able to solve for the acceleration of the \(\displaystyle 10kg\) crate.

 \(\displaystyle 35N=(10kg)(\vec{a})\)

\(\displaystyle \frac{35N}{10kg}=\vec{a}\)

\(\displaystyle 3.5\frac{m}{s^2}=\vec{a}\)

Derek observes that the crate is acceleration at a rate of \(\displaystyle 2\frac{m}{s^2}\), rather than the expected \(\displaystyle 3.5\frac{m}{s^2}\). An outside force is acting upon it to slow the acceleration.

The equation for the net force on the object is: \(\displaystyle \vec{F}_{resultant}=\vec{F}_{Derek}+\vec{F}_{friction}\). We also know, from Newton's second law, that \(\displaystyle \vec{F}_{resultant}=m\vec{a}_{resultant}\), where the resultant force and acceleration are the values actually observed.

Plug in the information we've been given so far to find the force of friction.

\(\displaystyle \vec{F}_{resultant}=\vec{F}_{Derek}+\vec{F}_{friction}\)

\(\displaystyle m\vec{a}_{resultant}=(35N)+\vec{F}_{friction}\)

\(\displaystyle (10kg)(2\frac{m}{s^2})=(35N)+\vec{F}_{friction}\)

\(\displaystyle 20N=35N+\vec{F}_{friction}\)

Subtract \(\displaystyle 35N\) from both sides to find the force of friction.

\(\displaystyle -15N=\vec{F}_{friction}\)

Friction will be negative because it acts in the direction opposite to the force of Derek.

Newton's third law states that when one object exerts a force on a second object, the second object exerts a force equal in size, but opposite in direction to the first. That means that the force of the hammer on the nail and the nail on the hammer will be equal in size, but opposite in direction.

Since the hammer exerts \(\displaystyle 32N\) of force on the nail, the nail must exert \(\displaystyle -32N\)of force on the hammer.

For the net force, we add up all the forces: \(\displaystyle F_{net}=F_{Michael}+F_{Annie}\).

Since force is a vector, the direction of the action matters. We will make leftward motion negative and rightward motion positive. Michael is pushing with \(\displaystyle 30N\) to the left, making his force equal to \(\displaystyle -30N\). Annie was pushing with \(\displaystyle 40N\) to the right, so her force will remain \(\displaystyle 40N\).

We can find the net force by adding the individual force together.

\(\displaystyle F_{net}=F_{Michael}+F_{Annie}\)

\(\displaystyle F_{net}=-30N+40N\)

\(\displaystyle F_{net}=10N\)

If the object has a constant velocity, that means that the net acceleration must be zero.

\(\displaystyle a=\frac{v_2-v_1}{t}\)

\(\displaystyle v_1=v_2\rightarrow a=\frac{0}{t}=0\)

In conjunction with Newton's second law, we can see that the net force is also zero. If there is no net acceleration, then there is no net force.

\(\displaystyle F_{net}=ma_{net}=m(0\frac{m}{s^2})=0N\)

Since Franklin is lifting the weight vertically, that means there will be two force acting upon the weight: his lifting force and gravity. The net force will be equal to the sum of the forces acting on the weight.

\(\displaystyle F_{net}=F_{lift}+F_{gravity}\)

Since we just proved that the net force will equal zero, we can say \(\displaystyle -F_{gravity}=F_{lift}\).

We know the mass of the weight and we know the acceleration, so we can solve for the lifting force.

\(\displaystyle -F_{gravity}=F_{lift}\)

\(\displaystyle -(mg)=F_{lift}\)

\(\displaystyle -(-9.8\frac{m}{s^2}*2kg)=F_{lift}\)

\(\displaystyle -(-19.6N)=F_{lift}\)

\(\displaystyle 19.6N=F_{lift}\)

The formula for force is \(\displaystyle F=ma\).

We are given the mass, but we will need to calculate the acceleration to use in the formula.

We know the initial velocity (zero because the box starts from rest), final velocity, and distance traveled. Using these values, we can find the acceleration using the formula \(\displaystyle v_f^2=v_i^2+2a\Delta x\).

Plug in our given values and solve for acceleration.

\(\displaystyle v_f^2=v_i^2+2a\Delta x\)

\(\displaystyle (3.45\frac{m}{s})^2=(0\frac{m}{s})^2+2a(12m)\)

\(\displaystyle 11.9\frac{m^2}{s^2}=a(24m)\)

Divide both sides by \(\displaystyle \small 24m\).

\(\displaystyle \frac{11.9\frac{m^2}{s^2}}{24m}=\frac{a( 24m)}{24m}{}\)

\(\displaystyle 0.496\frac{m}{s^2}=a\)

Now we know both the acceleration and the mass, allowing us to solve for the force.

\(\displaystyle F=ma\)

\(\displaystyle F=12kg* 0.496\frac{m}{s^2}\)

\(\displaystyle F=5.95N\)

In this problem there will be two forces acting upon the airplane: the weight of the plane (force of gravity) and the lifting force. Since we are looking for the minimum force to lift the plane, we can set the two forces equal to each other: \(\displaystyle F_{lift}=-F_g\).

We can calculate the gravitational force using the mass.

\(\displaystyle F_g=mg\)

\(\displaystyle F_g=94000kg*-9.8\frac{m}{s^2}\)

\(\displaystyle F_g=-921200N\)

Returning to the original equation, we see that the lifting force must be \(\displaystyle 921200N\).

\(\displaystyle F_{lift}=-F_g\)

\(\displaystyle F_{lift}=-(-921200N)=921200N\)

The formula for force is \(\displaystyle F=ma\). Plug in our given values and solve:

\(\displaystyle F=3kg*1.1\frac{m}{s^2}\)

\(\displaystyle F=3.3N\)