The mass of the truck is irrelevant to this problem.

The trailer has a mass of \(\displaystyle 2000kg\), and is accelerating at \(\displaystyle 1\frac{m}{s^2}\), thus, using

\(\displaystyle F=ma\)

\(\displaystyle F=2000*1\)

\(\displaystyle F=2000N\)

The mass of the truck is irrelevant to this problem.

The trailer has a mass of \(\displaystyle 2000kg\), and is accelerating at \(\displaystyle 1\frac{m}{s^2}\).

\(\displaystyle F=ma\)

\(\displaystyle F=2000*1\)

\(\displaystyle F=2000N\)

This is the force the truck is applying to the trailer to accelerate it forward. Based on Newton's Laws, the trailer is applying the same force, in the opposite direction.

We begin by drawing a free body diagram of the block.

\(\displaystyle F\) is the force applied to the block.

\(\displaystyle W\) is the weight of the block, or the force due to gravity. Weight is defined as \(\displaystyle W=mg\) where \(\displaystyle m\) is the mass of the block and \(\displaystyle g\) is the gravitational constant.

\(\displaystyle F_N\) is the normal force acting perpendicular to the contact surface.

Since the block is on ice, there is no friction.

We now apply Newton's second law in the direction of acceleration. In this problem, that is the x-direction.

Newton's second law is \(\displaystyle F_{net}=ma\) where \(\displaystyle F_{net}\) is the net force applied in the direction of acceleration, \(\displaystyle m\) is the mass of the block and \(\displaystyle a\) is the acceleration.

Solving for mass, we have

\(\displaystyle m=\frac{F_{net}}{a}\)

The only force in the direction of acceleration is the applied force \(\displaystyle F\), therefore

\(\displaystyle F_{net}=F=32N\). The problem tells us \(\displaystyle a=8\frac{m}{s^2}\). Substituting in this information gives us

\(\displaystyle m=\frac{F_{net}}{a}=\frac{32}{8}=4kg\)

We begin by drawing a free body diagram of the block.

\(\displaystyle F\) is the force applied to the block.

\(\displaystyle W\) is the weight of the block, or the force due to gravity. Weight is defined as \(\displaystyle W=mg\) where \(\displaystyle m\) is the mass of the block and \(\displaystyle g\) is the gravitational constant.

\(\displaystyle F_N\) is the normal force acting perpendicular to the contact surface.

Since the block is on ice, there is no friction.

We now apply Newton's second law in the direction of acceleration. In this problem, that is the x-direction.

Newton's second law is \(\displaystyle F_{net}=ma\) where \(\displaystyle F_{net}\) is the net force applied in the direction of acceleration, \(\displaystyle m\) is the mass of the block and \(\displaystyle a\) is the acceleration.

Solving for acceleration, we have

\(\displaystyle a=\frac{F_{net}}{m}\)

The only force in the direction of acceleration is the applied force \(\displaystyle F\), therefore

\(\displaystyle F_{net}=F=32N\). The problem tells us \(\displaystyle m=10kg\). Substituting in this information gives us

\(\displaystyle a=\frac{F_{net}}{m}=\frac{32}{10}=3.2\frac{m}{s^2}\)

We begin by drawing a free body diagram of the block.

\(\displaystyle F\) is the force applied to the block.

\(\displaystyle W\) is the weight of the block, or the force due to gravity. Weight is defined as \(\displaystyle W=mg\) where \(\displaystyle m\) is the mass of the block and \(\displaystyle g\) is the gravitational constant.

\(\displaystyle F_N\) is the normal force acting perpendicular to the contact surface.

Since the block is on ice, there is no friction. 

We now apply Newton's second law in the direction of acceleration.  In this problem, that is the x-direction.

Newton's second law is \(\displaystyle F_{net}=ma\) where \(\displaystyle F_{net}\) is the net force applied in the direction of acceleration, \(\displaystyle m\) is the mass of the block and \(\displaystyle a\) is the acceleration.

Solving for acceleration, we have

\(\displaystyle a=\frac{F_{net}}{m}\)

The only force in the direction of acceleration is the applied force \(\displaystyle F\), therefore

\(\displaystyle F_{net}=F=105N\).  The problem tells us \(\displaystyle m=45kg\).  Substituting in this information gives us

\(\displaystyle a=\frac{F_{net}}{m}=\frac{105}{45}=2.33\frac{m}{s^2}\)

We begin by drawing a free body diagram of the block.

\(\displaystyle F\) is the force applied to the block.

\(\displaystyle W\) is the weight of the block, or the force due to gravity. Weight is defined as \(\displaystyle W=mg\) where \(\displaystyle m\) is the mass of the block and \(\displaystyle g\) is the gravitational constant.

\(\displaystyle F_N\) is the normal force acting perpendicular to the contact surface.

Since the block is on ice, there is no friction.

We now apply Newton's second law in the direction of acceleration. In this problem, that is the x-direction.

Newton's second law is \(\displaystyle F_{net}=ma\) where \(\displaystyle F_{net}\) is the net force applied in the direction of acceleration, \(\displaystyle m\) is the mass of the block and \(\displaystyle a\) is the acceleration.

Solving for mass, we have

\(\displaystyle m=\frac{F_{net}}{a}\)

The only force in the direction of acceleration is the applied force \(\displaystyle F\), therefore

\(\displaystyle F_{net}=F=64N\).  The problem tells us \(\displaystyle a=4\frac{m}{s^2}\).  Substituting in this information gives us

\(\displaystyle m=\frac{F_{net}}{a}=\frac{64}{4}=16kg\)

We begin by drawing a free body diagram of the block.

\(\displaystyle F\) is the force applied to the block.

\(\displaystyle W\) is the weight of the block, or the force due to gravity. Weight is defined as \(\displaystyle W=mg\) where \(\displaystyle m\) is the mass of the block and \(\displaystyle g\) is the gravitational constant.

\(\displaystyle F_N\) is the normal force acting perpendicular to the contact surface.

Since the block is on ice, there is no friction.

We now apply Newton's second law in the direction of acceleration.  In this problem, that is the x-direction.

Newton's second law is \(\displaystyle F_{net}=ma\).

Where \(\displaystyle F_{net}\) is the net force applied in the direction of acceleration, \(\displaystyle m\) is the mass of the block and \(\displaystyle a\) is the acceleration.

The only force in the direction of acceleration is the applied force \(\displaystyle F\), therefore \(\displaystyle F_{net}=F\).  The problem tells us \(\displaystyle m=10kg\) and \(\displaystyle a=7\frac{m}{s^2}\). Substituting in this information gives us:

\(\displaystyle F=ma=10*7=70N\)

We begin by drawing a free body diagram of the block.

\(\displaystyle F\) is the force applied to the block.

\(\displaystyle W\) is the weight of the block, or the force due to gravity. Weight is defined as \(\displaystyle W=mg\) where \(\displaystyle m\) is the mass of the block and \(\displaystyle g\) is the gravitational constant.

\(\displaystyle F_N\) is the normal force acting perpendicular to the contact surface.

Since the block is on ice, there is no friction.

We now apply Newton's second law in the direction of acceleration. In this problem, that is the x-direction.

Newton's second law is \(\displaystyle F_{net}=ma\).

Where \(\displaystyle F_{net}\) is the net force applied in the direction of acceleration, \(\displaystyle m\) is the mass of the block and \(\displaystyle a\) is the acceleration.

The only force in the direction of acceleration is the applied force \(\displaystyle F\), therefore \(\displaystyle F_{net}=F\). The problem tells us \(\displaystyle m=8kg\) and \(\displaystyle a=3\frac{m}{s^2}\). Substituting in this information gives us

\(\displaystyle F=ma=8*3=24N\)

We begin by drawing a free body diagram of the block.

\(\displaystyle F\) is the force applied to the block.

\(\displaystyle W\) is the weight of the block, or the force due to gravity. Weight is defined as \(\displaystyle W=mg\) where \(\displaystyle m\) is the mass of the block and \(\displaystyle g\) is the gravitational constant.

\(\displaystyle F_N\) is the normal force acting perpendicular to the contact surface.

\(\displaystyle f\) is the force due to kinetic friction.  Friction is defined as \(\displaystyle f=\mu *F_N\) where \(\displaystyle \mu\) is the coefficient of friction.

To find the force due to friction, we need to find \(\displaystyle F_N\) by applying Newton's second in the y-direction. 

Newton's second law is \(\displaystyle F_{net}=ma\) where \(\displaystyle F_{net}\) is the net force, \(\displaystyle m\) is the mass of the block and \(\displaystyle a\) is the acceleration.

There are two forces in the y-direction, \(\displaystyle F_N\) and \(\displaystyle W\). There are in opposite directions, so they are subtracted. We are given \(\displaystyle m=10kg\). There is no acceleration in the y-direction, so \(\displaystyle a=0\).

Substituting all this information into Newton's second law gives us

\(\displaystyle F_{net}=ma\Rightarrow F_N-W=(10)*0\Rightarrow F_N-mg=(10)*0\)

\(\displaystyle F_N-mg=0\Rightarrow F_N=mg\)

Assuming \(\displaystyle g=10\), 

\(\displaystyle F_N=mg=(10)(10)=100N\).

Now that we have \(\displaystyle F_N\), we can find the force due to friction. Given that \(\displaystyle \mu=0.5\) and \(\displaystyle F_N=100N\),

\(\displaystyle f=\mu *F_N=0.5*100N=50N\)

We now apply Newton's second law in the direction of acceleration. In this problem, that is the x-direction.

Solving for acceleration, we have

\(\displaystyle a=\frac{F_{net}}{m}\)

There are two forces in the direction of acceleration,the applied force \(\displaystyle F\) and the force due to friction \(\displaystyle f\). Assuming that \(\displaystyle F\) is applied in the direction of acceleration and \(\displaystyle f\) is in the opposite direction,

\(\displaystyle F_{net}=F-f=132N-50N=82N\). The problem tells us \(\displaystyle m=10kg\).  Substituting in this information gives us

\(\displaystyle a=\frac{F_{net}}{m}=\frac{82}{10}=8.2\frac{m}{s^2}\)

We begin by drawing a free body diagram of the block.

\(\displaystyle F\) is the force applied to the block.

\(\displaystyle W\) is the weight of the block, or the force due to gravity. Weight is defined as \(\displaystyle W=mg\) where \(\displaystyle m\) is the mass of the block and \(\displaystyle g\) is the gravitational constant.

\(\displaystyle F_N\) is the normal force acting perpendicular to the contact surface.

\(\displaystyle f\) is the force due to kinetic friction.  Friction is defined as \(\displaystyle f=\mu *F_N\) where \(\displaystyle \mu\) is the coefficient of friction.

To find the force due to friction, we need to find \(\displaystyle F_N\) by applying Newton's second in the y-direction. 

Newton's second law is \(\displaystyle F_{net}=ma\) where \(\displaystyle F_{net}\) is the net force, \(\displaystyle m\) is the mass of the block and \(\displaystyle a\) is the acceleration.

There are two forces in the y-direction, \(\displaystyle F_N\) and \(\displaystyle W\). There are in opposite directions, so they are subtracted. We are given \(\displaystyle m=30kg\). There is no acceleration in the y-direction, so \(\displaystyle a=0\).

Substituting all this information into Newton's second law gives us:

\(\displaystyle F_{net}=ma\Rightarrow F_N-W=(30)*0\Rightarrow F_N-mg=(30)*0\)

\(\displaystyle F_N-mg=0\Rightarrow F_N=mg\)

Assuming \(\displaystyle g=10\)

\(\displaystyle F_N=mg=(30)(10)=300N\)

Now that we have \(\displaystyle F_N\), we can find the force due to friction. Given that \(\displaystyle \mu=0.2\) and \(\displaystyle F_N=300N\),

\(\displaystyle f=\mu *F_N=0.2*300N=60N\)

We now apply Newton's second law in the direction of acceleration. In this problem, that is the x-direction.

Solving for acceleration, we have

\(\displaystyle a=\frac{F_{net}}{m}\)

There are two forces in the direction of acceleration,the applied force \(\displaystyle F\) and the force due to friction \(\displaystyle f\).  Assuming that \(\displaystyle F\) is applied in the direction of acceleration and \(\displaystyle f\) is in the opposite direction,

\(\displaystyle F_{net}=F-f=90N-60N=30N\). The problem tells us \(\displaystyle m=30kg\). Substituting in this information gives us

\(\displaystyle a=\frac{F_{net}}{m}=\frac{30}{30}=1\frac{m}{s^2}\)