Weight is a particular force that is equal to mass times acceleration due to gravity. Start with Newton's second law,  .

Plug in the given values to solve.

Mass is a measure of how much matter is in an object, while weight is a measurement of the effective force of gravity on the object.

The amount of matter in the astronaut does not change; therefore, she has the same mass on the moon as she has on Earth.

Her weight, however, will change due to the change in gravity. The force of gravity will be the product of the acceleration and the astronaut's mass.

If the gravity is less on the moon than on Earth, then the force of gravity on the astronaut will also be less on the moon; thus, she will weigh less.

To determine the net force on a system, it is important to consider all the forces between each object in the horizontal direction.  The system is considered to be all the blocks together as they will all move with the same acceleration.

According to Newton’s 3rd Law  and  are equal in magnitude and opposite in direction so they will cancel out.  The same goes for  and .

Therefore our equation is reduce to 

From Newton’s 2nd law we know that the net force on a system is equal to the mass of the system times the acceleration of the system

Therefore

This means that the acceleration on any one block is

So when solving for the net force on any one block, the net force on the single block will be equal to that block's mass times the acceleration of the system.  Therefore for block B, the net force is

Consider the net forces acting on the object causing it to accelerate.

To determine the Force of Gravity in the x-direction, we must break the force of gravity into components and examine the side acting in the x-direction.  Using trigonometric functions we get that 

We know that the force of gravity is equal to mg

According to Newton’s 2nd law the force is equal to the mass times the acceleration of the object.

The force of friction is directly related to   (the coefficient of friction)  times the normal force.  In this case the normal force is equal to the y component of the force of gravity.

Therefore 

If we substitute this in our original net force equation

Notice that mass is in each piece of the equation so we can cancel it out.

We can also use our kinematic equations to determine the speed of the object at the bottom of the incline.  We can represent  to be the length of the slope.

Since the ball is assumed to be at rest at the top of the incline, the initial velocity at the top will be 0.

Therefore 

Solving for  on its own we get

In the original situation, the force of gravity is the only force pulling on the object.  Therefore the acceleration is 

We can substitute this value into our velocity equation

In the second situation the velocity is one half of the original velocity.  Therefore

We can place this back into our kinematic equation and rearrange it to solve for the acceleration.

We can then substitute that into our net force acceleration equation

Notice that g is in all of these terms so we can cancel it out.

Now we can substitute and solve

The equation for the force due to friction is , where μ is the coefficient of kinetic friction. Since there is only one force acting upon the object, the force due to friction, we can find its value using the equation . We can equate these two force equations, meaning that . We can solve for the normal force, but we need to find ma in order to find .

The problem gives us the mass of the crate, but we have to solve for the acceleration.

Start by finding the initial velocity. The problem gives us distance, final velocity, and change in time. We can use these values in the equation below to solve for the initial velocity.

Plug in our given values and solve.

We can use a linear motion equation to solve for the acceleration, using the velocity we just found. We now have the distance, time, and initial velocity.

Plug in the given values to solve for acceleration.

Now that we have the acceleration and the mass, we can return to our first equation for force.

The normal force is the same as the mass times gravity.

In this format, the masses cancel on both sides of the equation/

Now we can plug in our value for acceleration and gravity to solve for the coefficient of friction.

Consider the net forces acting on the object causing it to accelerate.

To determine the Force of Gravity in the x-direction, we must break the force of gravity into components and examine the side acting in the x-direction.  Using trigonometric functions we get that 

We know that the force of gravity is equal to mg

According to Newton’s 2nd law the force is equal to the mass times the acceleration of the object.

The force of friction is directly related to  μ (the coefficient of friction)  times the normal force.  In this case the normal force is equal to the y component of the force of gravity.

Therefore 

If we substitute this in our original net force equation

Notice that mass is in each piece of the equation so we can cancel it out.

Now we can rearrange and solve for the coefficient of friction

Since the only force pulling both cars down the slope is the force of gravity, both cars will accelerate at the same rate.

It is possible to prove this mathematically by examining the x-component of the gravitational pull.

We know that the force of gravity is equal to mg

According to Newton’s 2nd law the force is equal to the mass times the acceleration of the object.

Both sides have mass so this cancels out of the equation.

Since this is independent of mass, this will be true for both objects.

According to Newton’s 2nd Law the net force on the object is equal to the mass times the acceleration.  The net force is also equal to the sum of the forces involved.

In the second experiment the horizontal applied force is doubled.  However, this does not have any effect on the friction on the object.  Therefore the new equation is

From this equation we can see that the new acceleration will be more than double the original value.

To examine this numerically, let’s assume that the mass of the box is , the friction force is  and the applied force originally was .

Then, let’s double the horizontal force, keeping the friction force the same.

This value is more than double the original value, confirming our answer.

To determine the net force on a system, it is important to consider all the forces between each object in the same direction.  The direction to be considered is along the line of the rope.  The system is considered to be all the blocks together as they will all move with the same acceleration.

According to Newton’s 3rd Law, the Tension in the rope caused by A pulling on B and B pulling on A is equal in magnitude and opposite in direction.  Therefore these two Tension forces cancel out.

From Newton’s 2nd law we know that the net force on a system is equal to the mass of the system times the acceleration of the system

Therefore

We can calculate the force of gravity from the equation Fg = mg

Therefore

We can rearrange this equation and solve for a

The equation for the force of friction is , where μ is the coefficient of static friction.

The normal force is equal to the mass times acceleration due to gravity, but in the opposite direction (negative of the force of gravity).

Since the problem tells us that the force due to friction is , we can plug these values into our original equation to solve for the coefficient of friction.

The coefficient of friction has no units.