## Solution:

From Newton’s second law of motion, we know that

$\displaystyle \text{net F}=\text{ma}$

Considering the horizontal direction and let f be the resisting force, the equation becomes

$\displaystyle \text{F}-\text{f}=\text{ma}$

We are solving for F. Substituting the given values, we have

$\displaystyle \text{F}-400\:\text{N}=\left(245\:\text{kg}\right)\left(3.50\:\text{m/s}^2\right)$

$\displaystyle \text{F}=\left(245\:\text{kg}\right)\left(3.50\:\text{m/s}^2\right)+400\:\text{N}$

$\displaystyle \text{F}=857.5\:\text{N}+400\:\text{N}$

$\displaystyle \text{F}=1257.5\:\text{N}$        ◀

## Solution:

### Part A

If the acceleration of the child in the wagon is to be calculated, the system of interest is the wagon with the child in it.

### Part B

Considering all the forces acting on the wagon and child with mass m, there will be 5 forces. The horizontal forces are the two forces coming from the two other children, denoted as Fr and Fl. We also have friction going against the direction of motion.

The vertical forces are the weight of the wagon and the child and the normal force. These forces are shown in the free-body diagram shown.

### Part C

From Newton’s second law of motion, we know that

$\displaystyle \text{net F}=\text{ma}$

Considering the horizontal direction to the right as positive, the forces involved are Fr, Fl and f. Therefore, we have

$\displaystyle \text{F}_{\text{r}}-\text{F}_{\text{l}}-\text{f}=\text{ma}$

Solving for acceleration, we have

$\displaystyle \text{a}=\frac{\text{F}_\text{r}-\text{F}_\text{l}-\text{f}}{\text{m}}$

$\displaystyle \text{a}=\frac{90.0\:\text{N}-75.0\:\text{N}-12.0\:\text{N}}{23.0\:\text{kg}}$

$\displaystyle \text{a}=0.1304\:\text{m/s}^2$       ◀

### Part D

If the friction is 15 N, the acceleration would be

$\displaystyle \text{a}=\frac{90.0\:\text{N}-75.0\:\text{N}-15.0\:\text{N}}{23.0\:\text{kg}}$

$\displaystyle \text{a}=0\:\text{m/s}^2$       ◀

There would be no acceleration, meaning that the system is in equilibrium. The system is either not moving or it’s moving at a constant velocity.

## Solution:

We can answer this question by referring to the definition of acceleration–change in velocity over change in time. That is

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

Before substituting the given values, the units should be homogeneous. We will convert 1000 km/hr to m/s.

$\displaystyle 1000\:\frac{\text{km}}{\text{hr}}\times \frac{1000\:\text{m}}{1\:\text{km}}\times \frac{1\:\text{hr}}{3600\:\text{s}}=277.7778\:\text{m/s}$

We substitute the values into the equation,

$\displaystyle \text{a}=\frac{277.7778\:\text{m/s}}{1.1\:\text{s}}$

$\displaystyle \text{a}=252.5252\:\text{m/s}^2$       ◀

The deceleration is about 252.5252 m/s2

## Solution:

### Part A

If we look at the given figure, there are two forces responsible for the motion of the object in the horizontal direction–the thrust and the friction. We can state that the net force is equal to thrust minus friction. We are also given that T=25,900 Newtons. Based from Newton’s second law of motion, we can calculate for the acceleration as

$\displaystyle \text{net F}=\text{ma}$

$\displaystyle \text{a}=\frac{\text{net F}}{\text{m}}$

$\displaystyle \text{a}=\frac{2.4\times 10^4\:\text{N}-650\:\text{N}}{2100\:\text{kg}}$

$\displaystyle \text{a}=11.1180\:\text{m/s}^2$       ◀

### Part B

The magnitude of the acceleration depends on two forces–the thrust and the friction. Since the magnitude of the friction is the same no matter how many rockets are burning, then the acceleration can not be 1/4 when only 1 of the four rockets are burning.

## Solution:

From the problem, we know that the acceleration a is 196 m/s² and the mass is 2100 kg. We are asked to solve for the net force F. Basically, we can compute for the net force using the Second Law of Motion by Newton.

$\displaystyle \text{net F}=\text{ma}$

$\displaystyle \text{net F}=\left(2100\:\text{kg}\right)\left(196\:\text{m/s}^2\right)$

$\displaystyle \text{net F}=411,600\:\text{N}$       ◀

The net force opposing motion should be 411,600 Newtons to produce the given deceleration.

## Solution:

From the problem, we know that the net external force is 51 N. This net force is the difference between the force exerted by the man and the opposing friction force. That is

$\text{net F}=\text{F}-\text{f}$

So, we can solve for the force by the person to the mower.

$\text{F}=\text{net F}+\text{f}$

$\text{F}=51\:\text{N}\:+24\:\text{N}$

$\text{F}=75\:\text{N}$       ◀

We shall use kinematical equations to solve for the distance traveled by the mower after force F is removed. When F is removed, only f is now the only present force in the mower, so we can compute for the acceleration using the negative force.

$\displaystyle \text{a}=\frac{\text{F}}{\text{m}}$

$\displaystyle \text{a}=\frac{-24\:\text{N}}{24\:\text{kg}}$

$\text{a}=-1\:\text{m/s}^2$

We shall use this acceleration to compute for the distance travelled.

$\text{v}^2=\left(\text{v}_0\right)^2+2\text{ax}$

$\left(0\right)^2=\left(1.5\:\text{m/s}\right)^2+2\left(-1.0\:\text{m/s}^2\right)\text{x}$

$\text{x}=1.125\:\text{m}$       ◀

The mower will go 1.125 meters before stopping when the force F is already removed.

## Solution:

The velocity at time 2.5 seconds is 5 m/s.

The velocity at time 7.5 seconds is -4 m/s.