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.

\(\displaystyle Fnet = Fapplied + FAonB +FBonA + FBonC + FConB\)

According to Newton’s 3rd Law \(\displaystyle FAonB\) and \(\displaystyle FBonA\) are equal in magnitude and opposite in direction so they will cancel out.  The same goes for \(\displaystyle FBonC\) and \(\displaystyle FConB\).

Therefore our equation is reduce to 

\(\displaystyle Fnet = Fapplied\)

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

\(\displaystyle Fnet = (mA + mB +mC) *a\)

Therefore

\(\displaystyle Fapplied = (mA + mB +mC) *a\)

This means that the acceleration on any one block is

\(\displaystyle a = Fapplied/ (mA + mB +mC)\)

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

\(\displaystyle Fnet = mB * (Fapplied)/(mA+mB+mC)\)

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

\(\displaystyle Fnet = FGxdirection-Ffriction\)

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 

\(\displaystyle F_{GXdirection} = F_Gsin(\theta )\)

We know that the force of gravity is equal to mg

\(\displaystyle F_{GXdirection} = mgsin(\theta )\)

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

\(\displaystyle ma = mgsin(\theta )-F_{friction}\)

The force of friction is directly related to  \(\displaystyle \mu\) (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.

\(\displaystyle F_{GYdirection} = mgcos(\theta )\)

Therefore 

\(\displaystyle F_{friction}=\mu * F_N\)

\(\displaystyle F_N=F_{GYdirection}\)

\(\displaystyle F_{friction}=\mu *mgcos(\theta )\)

If we substitute this in our original net force equation

\(\displaystyle ma = mgsin(\theta )-\mu *mgcos(\theta )\)

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

\(\displaystyle a = gsin(\theta )-\mu gcos(\theta )\)

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

\(\displaystyle Vf2 = V02 + 2ax\)

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 

\(\displaystyle Vf2 = 2ax\)

Solving for \(\displaystyle Vf\) on its own we get

\(\displaystyle Vf=2ax\)

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

\(\displaystyle a = gsin(\theta )\)

We can substitute this value into our velocity equation

\(\displaystyle V_f=\sqrt{2gsin(\theta )x}\)

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

\(\displaystyle V_f=\frac{\sqrt{2gsin(\theta )x}}{2}\)

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

\(\displaystyle Vf2 = 2ax\)

\(\displaystyle (\frac{\sqrt{2gsin(\theta )x}}{2})^2= 2ax\)

\(\displaystyle \frac{2gsin(\theta )x}{4}=2ax\)

\(\displaystyle \frac{gsin(\theta )}{4}=a\)

We can then substitute that into our net force acceleration equation

\(\displaystyle \frac{gsin(\theta )}{4}= gsin(\theta )-\mu gcos(\theta )\)

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

\(\displaystyle \frac{sin(\theta )}{4}= sin(\theta )-\mu cos(\theta 30)\)

Now we can substitute and solve

\(\displaystyle \frac{sin(30)}{4}= sin(30)-\mu cos(30)\)

\(\displaystyle 0.125 = 0.5 -0.866 \mu\)

\(\displaystyle -.375 = -.866 \mu\)

\(\displaystyle \mu =0.43\)

The equation for the force due to friction is \(\displaystyle F_f=\mu FN\), 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 \(\displaystyle F=ma\). We can equate these two force equations, meaning that \(\displaystyle ma=\mu FN\). We can solve for the normal force, but we need to find ma in order to find \(\displaystyle \mu\).

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.

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

Plug in our given values and solve.

\(\displaystyle 10m=(0m/s+v_i)^2*5s\)

\(\displaystyle 10m/5s=v_i^2\)

\(\displaystyle 2m/s=v_i^2\)

\(\displaystyle 4m/s=v_i\)

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.

\(\displaystyle \Delta x=vit+1/2at^2\)

Plug in the given values to solve for acceleration.

\(\displaystyle 10m=(4ms)(5s)+1/2a(5s)^2\)

\(\displaystyle 10m=(20m)+1/2a(25s^2)\)

\(\displaystyle -10m=1/2a(25s^2)\)

\(\displaystyle -20m=a(25s^2)\)

\(\displaystyle -20m/25s^2=a\)

\(\displaystyle -0.8m/s^2=a\)

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

\(\displaystyle F=ma=\mu FN\)

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

\(\displaystyle ma= \mu (mg)\)

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

\(\displaystyle a= \mu *(g)\)

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

\(\displaystyle -0.8m/s^2=\mu (-9.8m/s^2)\)

\(\displaystyle -0.8m/s^2-9.8m/s^2= \mu\)

\(\displaystyle 0.082=\mu\)

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

\(\displaystyle F_{ne}t = F_{Gxdirection}-F_{friction}\)

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 

\(\displaystyle F_{GXdirection} = F_Gsin(\theta )\)

We know that the force of gravity is equal to mg

\(\displaystyle F_{GXdirection} = mgsin(\theta )\)

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

\(\displaystyle ma = mgsin(\theta )-F_{friction}\)

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.

\(\displaystyle F_{GYdirection} = mgcos(\theta )\)

Therefore 

\(\displaystyle F_{friction}=\mu *F_N\)

\(\displaystyle F_N=F_{GYdirection}\)

\(\displaystyle F_{friction}=\mu *mgcos(\theta )\)

If we substitute this in our original net force equation

\(\displaystyle ma = mgsin(\theta )-\mu *mgcos(\theta )\)

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

\(\displaystyle a = gsin(\theta )-\mu gcos(\theta )\)

Now we can rearrange and solve for the coefficient of friction

\(\displaystyle (.28m/s^2) = (9.8m/s^2)sin(25)- \mu (9.8m/s^2)cos(25)\)

\(\displaystyle .28 = 4.14 -8.88\mu\)

\(\displaystyle -3.86 = -8.88\mu\)

\(\displaystyle \mu =0.43\)

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.

\(\displaystyle F_{Net}=F_{GXdirection}\)

\(\displaystyle F_{GXdirection} = F_Gsin(\theta )\)

We know that the force of gravity is equal to mg

\(\displaystyle F_{GXdirection} = mgsin(\theta )\)

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

\(\displaystyle ma = mgsin(\theta )\)

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

\(\displaystyle a =gsin(\theta )\)

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.

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

\(\displaystyle F_{net} = F_{applied} - F_{friction}\)

\(\displaystyle ma=F_{applied}-F_{friction}\)

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

\(\displaystyle ma=2F_{applied}-F_{friction}\)

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 \(\displaystyle 1kg\), the friction force is \(\displaystyle 1N\) and the applied force originally was \(\displaystyle 2N\).

\(\displaystyle (1kg)*a=(2N-1N)\)

\(\displaystyle a=1m/s^2\)

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

\(\displaystyle (1kg)*a=(4N-1N)\)

\(\displaystyle a=3m/s^2\)

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.

\(\displaystyle Fnet = FTonA + FTonB + FGonB\)

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.

\(\displaystyle Fnet = FGonB\)

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

\(\displaystyle Fnet = (mA + mB)a\)

Therefore

\(\displaystyle FGonB = (mA +mB)a\)

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

\(\displaystyle FGonB = mB*g\)

Therefore

\(\displaystyle mB*g = (mA + mB)a\)

We can rearrange this equation and solve for a

\(\displaystyle a = mB*g/(mA +mB)\)

The equation for the force of friction is \(\displaystyle F_{friction}=\mu F_{normal}\), 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).

\(\displaystyle F_{normal}=-F_{gravity}=-(mg)\)

\(\displaystyle F_{normal}=-((12kg)(-9.8m/s^2))\)

\(\displaystyle F_{normal}=-(-117.6N)\)

\(\displaystyle F_{normal}=117.6N\)

Since the problem tells us that the force due to friction is \(\displaystyle -50N\), we can plug these values into our original equation to solve for the coefficient of friction.

\(\displaystyle F_{friction}=\mu F_{normal}\)

\(\displaystyle (-50N)=\mu (117.6N)\)

\(\displaystyle -50N/117.6N=\mu\)

\(\displaystyle 0.43=\mu\)

The coefficient of friction has no units.

Kinetic friction is never greater than static friction. More force is always requires to overcome static friction than is required to overcome kinetic friction. It can require a large force to initiate motion, causing an initial acceleration by overcoming static friction. Once motion has begun, however, less for is required to maintain the motion due to the principles of Newton's first law and inertia.