The mass of the truck is irrelevant to this problem.

The trailer has a mass of , and is accelerating at , thus, using

The mass of the truck is irrelevant to this problem.

The trailer has a mass of , and is accelerating at .

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.

 is the force applied to the block.

 is the weight of the block, or the force due to gravity. Weight is defined as  where  is the mass of the block and  is the gravitational constant.

 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  where  is the net force applied in the direction of acceleration,  is the mass of the block and  is the acceleration.

Solving for mass, we have

The only force in the direction of acceleration is the applied force , therefore

. The problem tells us . Substituting in this information gives us

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

 is the force applied to the block.

 is the weight of the block, or the force due to gravity. Weight is defined as  where  is the mass of the block and  is the gravitational constant.

 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  where  is the net force applied in the direction of acceleration,  is the mass of the block and  is the acceleration.

Solving for acceleration, we have

The only force in the direction of acceleration is the applied force , therefore

. The problem tells us . Substituting in this information gives us

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

 is the force applied to the block.

 is the weight of the block, or the force due to gravity. Weight is defined as  where  is the mass of the block and  is the gravitational constant.

 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  where  is the net force applied in the direction of acceleration,  is the mass of the block and  is the acceleration.

Solving for acceleration, we have

The only force in the direction of acceleration is the applied force , therefore

.  The problem tells us .  Substituting in this information gives us

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

 is the force applied to the block.

 is the weight of the block, or the force due to gravity. Weight is defined as  where  is the mass of the block and  is the gravitational constant.

 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  where  is the net force applied in the direction of acceleration,  is the mass of the block and  is the acceleration.

Solving for mass, we have

The only force in the direction of acceleration is the applied force , therefore

.  The problem tells us .  Substituting in this information gives us

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

 is the force applied to the block.

 is the weight of the block, or the force due to gravity. Weight is defined as  where  is the mass of the block and  is the gravitational constant.

 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 .

Where  is the net force applied in the direction of acceleration,  is the mass of the block and  is the acceleration.

The only force in the direction of acceleration is the applied force , therefore .  The problem tells us  and . Substituting in this information gives us:

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

 is the force applied to the block.

 is the weight of the block, or the force due to gravity. Weight is defined as  where  is the mass of the block and  is the gravitational constant.

 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 .

Where  is the net force applied in the direction of acceleration,  is the mass of the block and  is the acceleration.

The only force in the direction of acceleration is the applied force , therefore . The problem tells us  and . Substituting in this information gives us

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

 is the force applied to the block.

 is the weight of the block, or the force due to gravity. Weight is defined as  where  is the mass of the block and  is the gravitational constant.

 is the normal force acting perpendicular to the contact surface.

 is the force due to kinetic friction.  Friction is defined as  where  is the coefficient of friction.

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

Newton's second law is  where  is the net force,  is the mass of the block and  is the acceleration.

There are two forces in the y-direction,  and . There are in opposite directions, so they are subtracted. We are given . There is no acceleration in the y-direction, so .

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

Assuming , 

.

Now that we have , we can find the force due to friction. Given that  and ,

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

There are two forces in the direction of acceleration,the applied force  and the force due to friction . Assuming that  is applied in the direction of acceleration and  is in the opposite direction,

. The problem tells us .  Substituting in this information gives us

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

 is the force applied to the block.

 is the weight of the block, or the force due to gravity. Weight is defined as  where  is the mass of the block and  is the gravitational constant.

 is the normal force acting perpendicular to the contact surface.

 is the force due to kinetic friction.  Friction is defined as  where  is the coefficient of friction.

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

Newton's second law is  where  is the net force,  is the mass of the block and  is the acceleration.

There are two forces in the y-direction,  and . There are in opposite directions, so they are subtracted. We are given . There is no acceleration in the y-direction, so .

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

Assuming 

Now that we have , we can find the force due to friction. Given that  and ,

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

There are two forces in the direction of acceleration,the applied force  and the force due to friction .  Assuming that  is applied in the direction of acceleration and  is in the opposite direction,

. The problem tells us . Substituting in this information gives us