The correct answer is $\displaystyle 10.2\frac{m}{s^{2}}$. There are 2 forces acting on the box: tension force ($\displaystyle 200 N$) and normal force ($\displaystyle mass \cdot gravity$). First we must calculate the force of the box:

$\displaystyle F_{box}= (10.0kg)(9.8\frac{m}{s^{2}})= 98.0 N$

$\displaystyle F_{box} - mg = ma$ 

The equation can be manipulated to solve for acceleration:

$\displaystyle a = \frac{F_{box} - mg}{m}$

$\displaystyle a = \frac{200 N - 98.0 N}{10.0 kg} = 10.2 \frac{m}{s^{2}}$

^{$\displaystyle a=\frac{100N-98}{10kg}=0.20\frac{m}{s^2}$} 

Weight is the force that gravity exerts on an object. Weight is determined using the equation 

$\displaystyle W=mg$

Where $\displaystyle m$ is the mass of the object and $\displaystyle g$ is the gravitational constant on the given planet.

On Earth, $\displaystyle g=9.8\frac{m}{s^2}$, so the weight of the car on Earth is:

$\displaystyle W=mg=2700kg*9.8\frac{m}{s^2}= 26460N$

Weight is the force that gravity exerts on an object. Weight is determined using the equation 

$\displaystyle W=mg$

Where $\displaystyle m$ is the mass of the object and $\displaystyle g$ is the gravitational constant on the given planet.

On Mars, $\displaystyle g=3.75\frac{m}{s^2}$, so the weight of the car on Mars is:

$\displaystyle W=mg=2700kg*3.75\frac{m}{s^2}= 10125N$

To begin solving our problem, we will create a free body diagram, labeling all our forces and the direction in which they are acting. 

The normal force, $\displaystyle F_{n}$, is the component of force perpendicular to the surface of contact. Weight, $\displaystyle W$, is a force that gravity exerts on an object, acting in the same direction as gravity. To find the normal force, we will begin with Newton's 2nd law:

$\displaystyle \sum F=ma$ 

Where $\displaystyle \sum F$ is the sum of all forces in the same direction, $\displaystyle m$ is mass, and $\displaystyle a$ is the acceleration in the direction of the forces.

Since our object is at rest, $\displaystyle a=0$ and Newton's 2nd law becomes:

 $\displaystyle \sum F=0$

Assuming $\displaystyle F_{n}$ is in the positive direction, the sum of our forces in that direction is

$\displaystyle F_{n}-W=0$

Substituting in the definition of weight, $\displaystyle W=mg$, and solving the expression for $\displaystyle F_{n}$ gives us 

$\displaystyle F_{n}=mg$

Since $\displaystyle m=5kg$ and $\displaystyle g=9.8\frac{m}{s^2}$

$\displaystyle F_{n}=5*9.8=49N$

To begin solving our problem, we will create a free body diagram, labeling all our forces and the direction in which they are acting.

The normal force, $\displaystyle F_{n}$, is the component of force perpendicular to the surface of contact. Weight, $\displaystyle W$, is a force that gravity exerts on an object, acting in the same direction as gravity.

To find the normal force, we will begin with Newton's 2nd law:

$\displaystyle \sum F=ma$ 

Where $\displaystyle \sum F$ is the sum of all forces in the same direction, $\displaystyle m$ is mass, and $\displaystyle a$ is the acceleration in the direction of the forces.

Since our object is at rest, $\displaystyle a=0$ and Newton's 2nd law becomes

 $\displaystyle \sum F=0$

Assuming $\displaystyle F_{n}$ is in the positive direction, the sum of our forces in that direction is

$\displaystyle F_{n}-W=0$

Substituting in the definition of weight, $\displaystyle W=mg$, and solving the expression for $\displaystyle F_{n}$ gives us 

$\displaystyle F_{n}=mg$

Since $\displaystyle 15kg$ and $\displaystyle g=9.8\frac{m}{s^2}$

$\displaystyle F_{n}=15*9.8=147N$

To begin solving our problem, we will create a free body diagram, labeling all our forces and the direction in which they are acting. 

The normal force, $\displaystyle F_{n}$, is the component of force perpendicular to the surface of contact. Weight, $\displaystyle W$, is a force that gravity exerts on an object, acting in the same direction as gravity.

Weight must be divided into two components, the parts acting parallel and perpendicular to the contact surface.

$\displaystyle W_x$ is the x-component of the weight and is $\displaystyle W_x=Wsin\theta =mgsin\theta$

$\displaystyle W_y$ is the y-component of the weight and is $\displaystyle W_y=Wcos\theta =mgcos\theta$

To find the normal force, we will begin with Newton's 2nd law:

$\displaystyle \sum F=ma$ 

Where $\displaystyle \sum F$ is the sum of all forces in the same direction, $\displaystyle m$ is mass, and $\displaystyle a$ is the acceleration in the direction of the forces.

Since our object is at rest, $\displaystyle a=0$ and Newton's 2nd law becomes

 $\displaystyle \sum F=0$

To find $\displaystyle F_{n}$, we only need to consider forces acting the same or opposite direction as $\displaystyle F_{n}$.

Assuming $\displaystyle F_{n}$ is in the positive direction, the sum of our forces in that direction is

$\displaystyle F_{n}-W_y=0$

Substituting in $\displaystyle W_y$ , and solving the expression for $\displaystyle F_{n}$ gives us 

$\displaystyle F_n =mgcos\theta$

Since $\displaystyle 7.5kg$, $\displaystyle \theta=45^{\circ}$ and $\displaystyle g=9.8\frac{m}{s^2}$

$\displaystyle F_n =7.5*9.8cos(45) \approx 51.97N$

To begin solving our problem, we will create a free body diagram, labeling all our forces and the direction in which they are acting.

The normal force, $\displaystyle F_{n}$, is the component of force perpendicular to the surface of contact. Weight, $\displaystyle W$, is a force that gravity exerts on an object, acting in the same direction as gravity.

Weight must be divided into two components, the parts acting parallel and perpendicular to the contact surface.

$\displaystyle W_x$ is the x-component of the weight and is $\displaystyle W_x=Wsin\theta =mgsin\theta$

$\displaystyle W_y$ is the y-component of the weight and is $\displaystyle W_y=Wcos\theta =mgcos\theta$

To find the normal force, we will begin with Newton's 2nd law:

$\displaystyle \sum F=ma$ 

Where $\displaystyle \sum F$ is the sum of all forces in the same direction, $\displaystyle m$ is mass, and $\displaystyle a$ is the acceleration in the direction of the forces.

Since our object is at rest, $\displaystyle a=0$ and Newton's 2nd law becomes

 $\displaystyle \sum F=0$

To find $\displaystyle F_{n}$, we only need to consider forces acting the same or opposite direction as $\displaystyle F_{n}$.

Assuming $\displaystyle F_{n}$ is in the positive direction, the sum of our forces is

$\displaystyle F_{n}-W_y=0$

Substituting in $\displaystyle W_y$ , and solving the expression for $\displaystyle F_{n}$ gives us 

$\displaystyle F_n =mgcos\theta$

Since $\displaystyle m=10kg$, $\displaystyle \theta=30^{\circ}$ and $\displaystyle g=9.8\frac{m}{s^2}$

$\displaystyle F_n =10*9.8cos(30) \approx 84.87N$

To begin solving our problem, we will create a free body diagram, labeling all our forces and the direction in which they are acting.

The normal force, $\displaystyle F_{n}$, is the component of force perpendicular to the surface of contact. Weight, $\displaystyle W$, is a force that gravity exerts on an object, acting in the same direction as gravity. $\displaystyle F$ is the force pressing down on the book.

To find the normal force, we will begin with Newton's 2nd law:

$\displaystyle \sum F=ma$ 

Where $\displaystyle \sum F$ is the sum of all forces in the same direction, $\displaystyle m$ is mass, and $\displaystyle a$ is the acceleration in the direction of the forces.

Since our object is at rest, $\displaystyle a=0$ and Newton's 2nd law becomes

 $\displaystyle \sum F=0$

Assuming $\displaystyle F_{n}$ is in the positive direction, the sum of our forces in that direction is

$\displaystyle F_{n}-W-F=0$

Substituting in the definition of weight, $\displaystyle W=mg$, and solving the expression for $\displaystyle F_{n}$ gives us 

$\displaystyle F_{n}=mg+F$

Since $\displaystyle m=10kg$,  $\displaystyle F=10N$ and $\displaystyle g=9.8\frac{m}{s^2}$

$\displaystyle F_{n}=10*9.8+10=108N$

To begin solving our problem, we will create a free body diagram, labeling all our forces and the direction in which they are acting.

The normal force, $\displaystyle F_{n}$, is the component of force perpendicular to the surface of contact. Weight, $\displaystyle W$, is a force that gravity exerts on an object, acting in the same direction as gravity.  $\displaystyle F$ is the force pulling up on the book.

To find the normal force, we will begin with Newton's 2nd law:

$\displaystyle \sum F=ma$ 

Where $\displaystyle \sum F$ is the sum of all forces in the same direction, $\displaystyle m$ is mass, and $\displaystyle a$ is the acceleration in the direction of the forces.

Since our object is at rest, $\displaystyle a=0$ and Newton's 2nd law becomes

 $\displaystyle \sum F=0$

Assuming $\displaystyle F_{n}$ is in the positive direction, the sum of our forces acting in that direction is

$\displaystyle F_{n}+F-W=0$

Substituting in the definition of weight, $\displaystyle W=mg$, and solving the expression for $\displaystyle F_{n}$ gives us 

$\displaystyle F_{n}=mg-F$

Since $\displaystyle m=12kg$,  $\displaystyle F=15N$and $\displaystyle g=9.8\frac{m}{s^2}$

$\displaystyle F_{n}=12*9.8-15=102.6N$

Weight is the force that gravity exerts on an object. Weight is determined using the equation $\displaystyle W=mg$ where $\displaystyle m$ is the mass of the object and $\displaystyle g$ is the gravitational constant on the given planet.

Given the weight of an object and the gravitational constant acting on it, we find the mass using

$\displaystyle W=mg\Rightarrow m=\frac{W}{g}=\frac{60000N}{9.8\frac{m}{s^2}} \approx 6122kg$