The net force acting on the entire two-block system.

Question

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Answer 1

Question

Chapter 4, Problem 5SP

(a)

To determine

The net force acting on the entire two-block system.

(a)

Expert Solution

Answer to Problem 5SP

The net force acting on the entire two-block system is $16 N$ .

Explanation of Solution

Given info: The horizontal force acting on $4 kg$ mass along the right direction is $30 N$ , that along the left direction is $8 N$ and horizontal force acting on $2 kg$ mass along left direction is $6 N$ .

Write the expression for the net horizontal force.

$Fnet=Fright−Fleft$

Here,

$Fnet$ is the net force acting on the system

$Fright$ is the horizontal force acting in the right direction

$Fleft$ is the horizontal force acting in the left direction

Total force on the left direction is the sum of $6 N$ and $8 N$ .

That is,

$Fleft=6 N+8 N=14 N$

Substitute $14 N$ for $Fleft$ and $30 N$ for $Fright$ in the above equation to get $Fnet$ .

$Fnet=30 N−14 N=16 N$

Conclusion:

Thus, the net force acting on the entire two-block system is $16 N$ .

(b)

To determine

The acceleration of the system.

(b)

Expert Solution

Answer to Problem 5SP

The acceleration of the system is $2.67 m/s2$ .

Explanation of Solution

Given info: The masses of the blocks are $2 kg$ and $6 kg$ .

Write the expression for the acceleration of the horizontal acceleration of the block.

$a=Fnetm$

Here,

$m$ is the total mass of the system

$a$ is the horizontal acceleration of the system

Total mass of the system is $2 kg+4 kg=6 kg$ .

Substitute $16 N$ for $Fnet$ and $6 kg$ for $m$ in the above equation to get $a$ .

$a=16 N6 kg=2.67 m/s2$

Conclusion:

Thus, the acceleration of the system is $2.67 m/s2$ .

(c)

To determine

The force acting on the $2 kg$ mass by the connecting string.

(c)

Expert Solution

Answer to Problem 5SP

The force acting on the $2 kg$ mass by the connecting string is $15.6 N$ .

Explanation of Solution

Given info: The horizontal force acting on $2 kg$ mass along the left direction is $6 N$ .

Let $m1$ be the mass of $2 kg$ block and $m2$ be the mass of $4 kg$ block.

Write the expression for the net force on $2 kg$ mass.

$Fnet1=m1a$

Here,

$m1$ is the mass of $2 kg$ block

$Fnet1$ is the net horizontal force on the mass $2 kg$

Substitute $2 kg$ for $m1$ and $2.67 m/s2$ for $a$ in the above equation to get $Fnet1$ .

$Fnet1=(2 kg )(2.67 m/s2)=5.34 N$

This net force is acting along the direction of acceleration. That is, along the right direction.

Write the expression for the net force on the $2 kg$ mass.

$Fnet1=Ftension−Fleft$

Here,

$Fleft$ is the horizontal force on the left direction of mass $2 kg$

$Ftension$ is the tension on the string

The negative sign indicate that the force of tension is along the right direction whereas the $Fleft$ is along the left direction.

Substitute $6 N$ for $Fleft$ and $5.34 N$ for $Fnet1$ in the above equation to get $Ftension$ .

$5.34 N=Ftension−6 NFtension=11.34N$

Therefore, the force acting on the $2 kg$ by the string is equal to $11.34N$ .

Conclusion:

Thus, the force acting on the $3 kg$ mass by the connecting string is $11.34N$ .

(d)

To determine

The net force and acceleration of $4 kg$ block.

(d)

Expert Solution

Answer to Problem 5SP

The net force on the $4 kg$ block is $10.67 N$ and net acceleration of the block is $2.67 m/s2$ .

Explanation of Solution

Given info: The horizontal force acting on $4 kg$ mass along the right direction is $30 N$ and that along the left direction is $8 N$

Write the expression for the net force on the $4 kg$ mass.

$Fnet2=Fright−(Ftension+Fleft)$

Here,

$Fleft$ is the horizontal force on the left direction of mass $4 kg$

$Fright$ is the horizontal force on the right direction of mass $4 kg$

$Ftension$ is the tension on the string

Substitute $30 N$ for $Fright$ , $11.34 N$ for $Ftension$ and $8N$ for $Fleft$ in the above equation to get $Fnet2$ .

$Fnet2=30 N−(11.34 N+8N)=10.66 N$

Write the expression for the net acceleration on $4 kg$ mass.

$a=Fnet2m2$

Here,

$m2$ is the $4 kg$ mass

$a$ is the net acceleration

Substitute $4 kg$ for $m2$ and $10.66 N$ for $Fnet2$ in the above equation to get $Fnet2$ .

$a=10.66 N4 kg=2.67m/s2$

This is same as the net acceleration obtained in part $b$ .

Conclusion:

Thus, the net force on the $4 kg$ block is $10.66 N$ and net acceleration of the block is $2.67m/s2$ $3.2 m/s2$ .

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Answer 2

Question

Chapter 4, Problem 5SP

(a)

To determine

The net force acting on the entire two-block system.

(a)

Expert Solution

Answer to Problem 5SP

The net force acting on the entire two-block system is $16 N$ .

Explanation of Solution

Given info: The horizontal force acting on $4 kg$ mass along the right direction is $30 N$ , that along the left direction is $8 N$ and horizontal force acting on $2 kg$ mass along left direction is $6 N$ .

Write the expression for the net horizontal force.

$Fnet=Fright−Fleft$

Here,

$Fnet$ is the net force acting on the system

$Fright$ is the horizontal force acting in the right direction

$Fleft$ is the horizontal force acting in the left direction

Total force on the left direction is the sum of $6 N$ and $8 N$ .

That is,

$Fleft=6 N+8 N=14 N$

Substitute $14 N$ for $Fleft$ and $30 N$ for $Fright$ in the above equation to get $Fnet$ .

$Fnet=30 N−14 N=16 N$

Conclusion:

Thus, the net force acting on the entire two-block system is $16 N$ .

(b)

To determine

The acceleration of the system.

(b)

Expert Solution

Answer to Problem 5SP

The acceleration of the system is $2.67 m/s2$ .

Explanation of Solution

Given info: The masses of the blocks are $2 kg$ and $6 kg$ .

Write the expression for the acceleration of the horizontal acceleration of the block.

$a=Fnetm$

Here,

$m$ is the total mass of the system

$a$ is the horizontal acceleration of the system

Total mass of the system is $2 kg+4 kg=6 kg$ .

Substitute $16 N$ for $Fnet$ and $6 kg$ for $m$ in the above equation to get $a$ .

$a=16 N6 kg=2.67 m/s2$

Conclusion:

Thus, the acceleration of the system is $2.67 m/s2$ .

(c)

To determine

The force acting on the $2 kg$ mass by the connecting string.

(c)

Expert Solution

Answer to Problem 5SP

The force acting on the $2 kg$ mass by the connecting string is $15.6 N$ .

Explanation of Solution

Given info: The horizontal force acting on $2 kg$ mass along the left direction is $6 N$ .

Let $m1$ be the mass of $2 kg$ block and $m2$ be the mass of $4 kg$ block.

Write the expression for the net force on $2 kg$ mass.

$Fnet1=m1a$

Here,

$m1$ is the mass of $2 kg$ block

$Fnet1$ is the net horizontal force on the mass $2 kg$

Substitute $2 kg$ for $m1$ and $2.67 m/s2$ for $a$ in the above equation to get $Fnet1$ .

$Fnet1=(2 kg )(2.67 m/s2)=5.34 N$

This net force is acting along the direction of acceleration. That is, along the right direction.

Write the expression for the net force on the $2 kg$ mass.

$Fnet1=Ftension−Fleft$

Here,

$Fleft$ is the horizontal force on the left direction of mass $2 kg$

$Ftension$ is the tension on the string

The negative sign indicate that the force of tension is along the right direction whereas the $Fleft$ is along the left direction.

Substitute $6 N$ for $Fleft$ and $5.34 N$ for $Fnet1$ in the above equation to get $Ftension$ .

$5.34 N=Ftension−6 NFtension=11.34N$

Therefore, the force acting on the $2 kg$ by the string is equal to $11.34N$ .

Conclusion:

Thus, the force acting on the $3 kg$ mass by the connecting string is $11.34N$ .

(d)

To determine

The net force and acceleration of $4 kg$ block.

(d)

Expert Solution

Answer to Problem 5SP

The net force on the $4 kg$ block is $10.67 N$ and net acceleration of the block is $2.67 m/s2$ .

Explanation of Solution

Given info: The horizontal force acting on $4 kg$ mass along the right direction is $30 N$ and that along the left direction is $8 N$

Write the expression for the net force on the $4 kg$ mass.

$Fnet2=Fright−(Ftension+Fleft)$

Here,

$Fleft$ is the horizontal force on the left direction of mass $4 kg$

$Fright$ is the horizontal force on the right direction of mass $4 kg$

$Ftension$ is the tension on the string

Substitute $30 N$ for $Fright$ , $11.34 N$ for $Ftension$ and $8N$ for $Fleft$ in the above equation to get $Fnet2$ .

$Fnet2=30 N−(11.34 N+8N)=10.66 N$

Write the expression for the net acceleration on $4 kg$ mass.

$a=Fnet2m2$

Here,

$m2$ is the $4 kg$ mass

$a$ is the net acceleration

Substitute $4 kg$ for $m2$ and $10.66 N$ for $Fnet2$ in the above equation to get $Fnet2$ .

$a=10.66 N4 kg=2.67m/s2$

This is same as the net acceleration obtained in part $b$ .

Conclusion:

Thus, the net force on the $4 kg$ block is $10.66 N$ and net acceleration of the block is $2.67m/s2$ $3.2 m/s2$ .

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Answer 3

Question

Chapter 4, Problem 5SP

(a)

To determine

The net force acting on the entire two-block system.

(a)

Expert Solution

Answer to Problem 5SP

The net force acting on the entire two-block system is $16 N$ .

Explanation of Solution

Given info: The horizontal force acting on $4 kg$ mass along the right direction is $30 N$ , that along the left direction is $8 N$ and horizontal force acting on $2 kg$ mass along left direction is $6 N$ .

Write the expression for the net horizontal force.

$Fnet=Fright−Fleft$

Here,

$Fnet$ is the net force acting on the system

$Fright$ is the horizontal force acting in the right direction

$Fleft$ is the horizontal force acting in the left direction

Total force on the left direction is the sum of $6 N$ and $8 N$ .

That is,

$Fleft=6 N+8 N=14 N$

Substitute $14 N$ for $Fleft$ and $30 N$ for $Fright$ in the above equation to get $Fnet$ .

$Fnet=30 N−14 N=16 N$

Conclusion:

Thus, the net force acting on the entire two-block system is $16 N$ .

(b)

To determine

The acceleration of the system.

(b)

Expert Solution

Answer to Problem 5SP

The acceleration of the system is $2.67 m/s2$ .

Explanation of Solution

Given info: The masses of the blocks are $2 kg$ and $6 kg$ .

Write the expression for the acceleration of the horizontal acceleration of the block.

$a=Fnetm$

Here,

$m$ is the total mass of the system

$a$ is the horizontal acceleration of the system

Total mass of the system is $2 kg+4 kg=6 kg$ .

Substitute $16 N$ for $Fnet$ and $6 kg$ for $m$ in the above equation to get $a$ .

$a=16 N6 kg=2.67 m/s2$

Conclusion:

Thus, the acceleration of the system is $2.67 m/s2$ .

(c)

To determine

The force acting on the $2 kg$ mass by the connecting string.

(c)

Expert Solution

Answer to Problem 5SP

The force acting on the $2 kg$ mass by the connecting string is $15.6 N$ .

Explanation of Solution

Given info: The horizontal force acting on $2 kg$ mass along the left direction is $6 N$ .

Let $m1$ be the mass of $2 kg$ block and $m2$ be the mass of $4 kg$ block.

Write the expression for the net force on $2 kg$ mass.

$Fnet1=m1a$

Here,

$m1$ is the mass of $2 kg$ block

$Fnet1$ is the net horizontal force on the mass $2 kg$

Substitute $2 kg$ for $m1$ and $2.67 m/s2$ for $a$ in the above equation to get $Fnet1$ .

$Fnet1=(2 kg )(2.67 m/s2)=5.34 N$

This net force is acting along the direction of acceleration. That is, along the right direction.

Write the expression for the net force on the $2 kg$ mass.

$Fnet1=Ftension−Fleft$

Here,

$Fleft$ is the horizontal force on the left direction of mass $2 kg$

$Ftension$ is the tension on the string

The negative sign indicate that the force of tension is along the right direction whereas the $Fleft$ is along the left direction.

Substitute $6 N$ for $Fleft$ and $5.34 N$ for $Fnet1$ in the above equation to get $Ftension$ .

$5.34 N=Ftension−6 NFtension=11.34N$

Therefore, the force acting on the $2 kg$ by the string is equal to $11.34N$ .

Conclusion:

Thus, the force acting on the $3 kg$ mass by the connecting string is $11.34N$ .

(d)

To determine

The net force and acceleration of $4 kg$ block.

(d)

Expert Solution

Answer to Problem 5SP

The net force on the $4 kg$ block is $10.67 N$ and net acceleration of the block is $2.67 m/s2$ .

Explanation of Solution

Given info: The horizontal force acting on $4 kg$ mass along the right direction is $30 N$ and that along the left direction is $8 N$

Write the expression for the net force on the $4 kg$ mass.

$Fnet2=Fright−(Ftension+Fleft)$

Here,

$Fleft$ is the horizontal force on the left direction of mass $4 kg$

$Fright$ is the horizontal force on the right direction of mass $4 kg$

$Ftension$ is the tension on the string

Substitute $30 N$ for $Fright$ , $11.34 N$ for $Ftension$ and $8N$ for $Fleft$ in the above equation to get $Fnet2$ .

$Fnet2=30 N−(11.34 N+8N)=10.66 N$

Write the expression for the net acceleration on $4 kg$ mass.

$a=Fnet2m2$

Here,

$m2$ is the $4 kg$ mass

$a$ is the net acceleration

Substitute $4 kg$ for $m2$ and $10.66 N$ for $Fnet2$ in the above equation to get $Fnet2$ .

$a=10.66 N4 kg=2.67m/s2$

This is same as the net acceleration obtained in part $b$ .

Conclusion:

Thus, the net force on the $4 kg$ block is $10.66 N$ and net acceleration of the block is $2.67m/s2$ $3.2 m/s2$ .

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Answer 4

Question

Chapter 4, Problem 5SP

(a)

To determine

The net force acting on the entire two-block system.

(a)

Expert Solution

Answer to Problem 5SP

The net force acting on the entire two-block system is $16 N$ .

Explanation of Solution

Given info: The horizontal force acting on $4 kg$ mass along the right direction is $30 N$ , that along the left direction is $8 N$ and horizontal force acting on $2 kg$ mass along left direction is $6 N$ .

Write the expression for the net horizontal force.

$Fnet=Fright−Fleft$

Here,

$Fnet$ is the net force acting on the system

$Fright$ is the horizontal force acting in the right direction

$Fleft$ is the horizontal force acting in the left direction

Total force on the left direction is the sum of $6 N$ and $8 N$ .

That is,

$Fleft=6 N+8 N=14 N$

Substitute $14 N$ for $Fleft$ and $30 N$ for $Fright$ in the above equation to get $Fnet$ .

$Fnet=30 N−14 N=16 N$

Conclusion:

Thus, the net force acting on the entire two-block system is $16 N$ .

(b)

To determine

The acceleration of the system.

(b)

Expert Solution

Answer to Problem 5SP

The acceleration of the system is $2.67 m/s2$ .

Explanation of Solution

Given info: The masses of the blocks are $2 kg$ and $6 kg$ .

Write the expression for the acceleration of the horizontal acceleration of the block.

$a=Fnetm$

Here,

$m$ is the total mass of the system

$a$ is the horizontal acceleration of the system

Total mass of the system is $2 kg+4 kg=6 kg$ .

Substitute $16 N$ for $Fnet$ and $6 kg$ for $m$ in the above equation to get $a$ .

$a=16 N6 kg=2.67 m/s2$

Conclusion:

Thus, the acceleration of the system is $2.67 m/s2$ .

(c)

To determine

The force acting on the $2 kg$ mass by the connecting string.

(c)

Expert Solution

Answer to Problem 5SP

The force acting on the $2 kg$ mass by the connecting string is $15.6 N$ .

Explanation of Solution

Given info: The horizontal force acting on $2 kg$ mass along the left direction is $6 N$ .

Let $m1$ be the mass of $2 kg$ block and $m2$ be the mass of $4 kg$ block.

Write the expression for the net force on $2 kg$ mass.

$Fnet1=m1a$

Here,

$m1$ is the mass of $2 kg$ block

$Fnet1$ is the net horizontal force on the mass $2 kg$

Substitute $2 kg$ for $m1$ and $2.67 m/s2$ for $a$ in the above equation to get $Fnet1$ .

$Fnet1=(2 kg )(2.67 m/s2)=5.34 N$

This net force is acting along the direction of acceleration. That is, along the right direction.

Write the expression for the net force on the $2 kg$ mass.

$Fnet1=Ftension−Fleft$

Here,

$Fleft$ is the horizontal force on the left direction of mass $2 kg$

$Ftension$ is the tension on the string

The negative sign indicate that the force of tension is along the right direction whereas the $Fleft$ is along the left direction.

Substitute $6 N$ for $Fleft$ and $5.34 N$ for $Fnet1$ in the above equation to get $Ftension$ .

$5.34 N=Ftension−6 NFtension=11.34N$

Therefore, the force acting on the $2 kg$ by the string is equal to $11.34N$ .

Conclusion:

Thus, the force acting on the $3 kg$ mass by the connecting string is $11.34N$ .

(d)

To determine

The net force and acceleration of $4 kg$ block.

(d)

Expert Solution

Answer to Problem 5SP

The net force on the $4 kg$ block is $10.67 N$ and net acceleration of the block is $2.67 m/s2$ .

Explanation of Solution

Given info: The horizontal force acting on $4 kg$ mass along the right direction is $30 N$ and that along the left direction is $8 N$

Write the expression for the net force on the $4 kg$ mass.

$Fnet2=Fright−(Ftension+Fleft)$

Here,

$Fleft$ is the horizontal force on the left direction of mass $4 kg$

$Fright$ is the horizontal force on the right direction of mass $4 kg$

$Ftension$ is the tension on the string

Substitute $30 N$ for $Fright$ , $11.34 N$ for $Ftension$ and $8N$ for $Fleft$ in the above equation to get $Fnet2$ .

$Fnet2=30 N−(11.34 N+8N)=10.66 N$

Write the expression for the net acceleration on $4 kg$ mass.

$a=Fnet2m2$

Here,

$m2$ is the $4 kg$ mass

$a$ is the net acceleration

Substitute $4 kg$ for $m2$ and $10.66 N$ for $Fnet2$ in the above equation to get $Fnet2$ .

$a=10.66 N4 kg=2.67m/s2$

This is same as the net acceleration obtained in part $b$ .

Conclusion:

Thus, the net force on the $4 kg$ block is $10.66 N$ and net acceleration of the block is $2.67m/s2$ $3.2 m/s2$ .

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Answer 5

Chapter 4, Problem 5SP

(a)

To determine

The net force acting on the entire two-block system.

(a)

Expert Solution

Answer to Problem 5SP

The net force acting on the entire two-block system is $16 N$ .

Explanation of Solution

Given info: The horizontal force acting on $4 kg$ mass along the right direction is $30 N$ , that along the left direction is $8 N$ and horizontal force acting on $2 kg$ mass along left direction is $6 N$ .

Write the expression for the net horizontal force.

$Fnet=Fright−Fleft$

Here,

$Fnet$ is the net force acting on the system

$Fright$ is the horizontal force acting in the right direction

$Fleft$ is the horizontal force acting in the left direction

Total force on the left direction is the sum of $6 N$ and $8 N$ .

That is,

$Fleft=6 N+8 N=14 N$

Substitute $14 N$ for $Fleft$ and $30 N$ for $Fright$ in the above equation to get $Fnet$ .

$Fnet=30 N−14 N=16 N$

Conclusion:

Thus, the net force acting on the entire two-block system is $16 N$ .

(b)

To determine

The acceleration of the system.

(b)

Expert Solution

Answer to Problem 5SP

The acceleration of the system is $2.67 m/s2$ .

Explanation of Solution

Given info: The masses of the blocks are $2 kg$ and $6 kg$ .

Write the expression for the acceleration of the horizontal acceleration of the block.

$a=Fnetm$

Here,

$m$ is the total mass of the system

$a$ is the horizontal acceleration of the system

Total mass of the system is $2 kg+4 kg=6 kg$ .

Substitute $16 N$ for $Fnet$ and $6 kg$ for $m$ in the above equation to get $a$ .

$a=16 N6 kg=2.67 m/s2$

Conclusion:

Thus, the acceleration of the system is $2.67 m/s2$ .

(c)

To determine

The force acting on the $2 kg$ mass by the connecting string.

(c)

Expert Solution

Answer to Problem 5SP

The force acting on the $2 kg$ mass by the connecting string is $15.6 N$ .

Explanation of Solution

Given info: The horizontal force acting on $2 kg$ mass along the left direction is $6 N$ .

Let $m1$ be the mass of $2 kg$ block and $m2$ be the mass of $4 kg$ block.

Write the expression for the net force on $2 kg$ mass.

$Fnet1=m1a$

Here,

$m1$ is the mass of $2 kg$ block

$Fnet1$ is the net horizontal force on the mass $2 kg$

Substitute $2 kg$ for $m1$ and $2.67 m/s2$ for $a$ in the above equation to get $Fnet1$ .

$Fnet1=(2 kg )(2.67 m/s2)=5.34 N$

This net force is acting along the direction of acceleration. That is, along the right direction.

Write the expression for the net force on the $2 kg$ mass.

$Fnet1=Ftension−Fleft$

Here,

$Fleft$ is the horizontal force on the left direction of mass $2 kg$

$Ftension$ is the tension on the string

The negative sign indicate that the force of tension is along the right direction whereas the $Fleft$ is along the left direction.

Substitute $6 N$ for $Fleft$ and $5.34 N$ for $Fnet1$ in the above equation to get $Ftension$ .

$5.34 N=Ftension−6 NFtension=11.34N$

Therefore, the force acting on the $2 kg$ by the string is equal to $11.34N$ .

Conclusion:

Thus, the force acting on the $3 kg$ mass by the connecting string is $11.34N$ .

(d)

To determine

The net force and acceleration of $4 kg$ block.

(d)

Expert Solution

Answer to Problem 5SP

The net force on the $4 kg$ block is $10.67 N$ and net acceleration of the block is $2.67 m/s2$ .

Explanation of Solution

Given info: The horizontal force acting on $4 kg$ mass along the right direction is $30 N$ and that along the left direction is $8 N$

Write the expression for the net force on the $4 kg$ mass.

$Fnet2=Fright−(Ftension+Fleft)$

Here,

$Fleft$ is the horizontal force on the left direction of mass $4 kg$

$Fright$ is the horizontal force on the right direction of mass $4 kg$

$Ftension$ is the tension on the string

Substitute $30 N$ for $Fright$ , $11.34 N$ for $Ftension$ and $8N$ for $Fleft$ in the above equation to get $Fnet2$ .

$Fnet2=30 N−(11.34 N+8N)=10.66 N$

Write the expression for the net acceleration on $4 kg$ mass.

$a=Fnet2m2$

Here,

$m2$ is the $4 kg$ mass

$a$ is the net acceleration

Substitute $4 kg$ for $m2$ and $10.66 N$ for $Fnet2$ in the above equation to get $Fnet2$ .

$a=10.66 N4 kg=2.67m/s2$

This is same as the net acceleration obtained in part $b$ .

Conclusion:

Thus, the net force on the $4 kg$ block is $10.66 N$ and net acceleration of the block is $2.67m/s2$ $3.2 m/s2$ .

Answer 6

Chapter 4, Problem 5SP

(a)

To determine

The net force acting on the entire two-block system.

(a)

Expert Solution

Answer to Problem 5SP

The net force acting on the entire two-block system is $16 N$ .

Explanation of Solution

Given info: The horizontal force acting on $4 kg$ mass along the right direction is $30 N$ , that along the left direction is $8 N$ and horizontal force acting on $2 kg$ mass along left direction is $6 N$ .

Write the expression for the net horizontal force.

$Fnet=Fright−Fleft$

Here,

$Fnet$ is the net force acting on the system

$Fright$ is the horizontal force acting in the right direction

$Fleft$ is the horizontal force acting in the left direction

Total force on the left direction is the sum of $6 N$ and $8 N$ .

That is,

$Fleft=6 N+8 N=14 N$

Substitute $14 N$ for $Fleft$ and $30 N$ for $Fright$ in the above equation to get $Fnet$ .

$Fnet=30 N−14 N=16 N$

Conclusion:

Thus, the net force acting on the entire two-block system is $16 N$ .

Answer 7

Chapter 4, Problem 5SP

(a)

To determine

The net force acting on the entire two-block system.

Answer 8

(a)

Expert Solution

Answer to Problem 5SP

The net force acting on the entire two-block system is $16 N$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given info: The horizontal force acting on $4 kg$ mass along the right direction is $30 N$ , that along the left direction is $8 N$ and horizontal force acting on $2 kg$ mass along left direction is $6 N$ .

Write the expression for the net horizontal force.

$Fnet=Fright−Fleft$

Here,

$Fnet$ is the net force acting on the system

$Fright$ is the horizontal force acting in the right direction

$Fleft$ is the horizontal force acting in the left direction

Total force on the left direction is the sum of $6 N$ and $8 N$ .

That is,

$Fleft=6 N+8 N=14 N$

Substitute $14 N$ for $Fleft$ and $30 N$ for $Fright$ in the above equation to get $Fnet$ .

$Fnet=30 N−14 N=16 N$

Conclusion:

Thus, the net force acting on the entire two-block system is $16 N$ .

Answer 13

(b)

To determine

The acceleration of the system.

(b)

Expert Solution

Answer to Problem 5SP

The acceleration of the system is $2.67 m/s2$ .

Explanation of Solution

Given info: The masses of the blocks are $2 kg$ and $6 kg$ .

Write the expression for the acceleration of the horizontal acceleration of the block.

$a=Fnetm$

Here,

$m$ is the total mass of the system

$a$ is the horizontal acceleration of the system

Total mass of the system is $2 kg+4 kg=6 kg$ .

Substitute $16 N$ for $Fnet$ and $6 kg$ for $m$ in the above equation to get $a$ .

$a=16 N6 kg=2.67 m/s2$

Conclusion:

Thus, the acceleration of the system is $2.67 m/s2$ .

Answer 14

(b)

To determine

The acceleration of the system.

Answer 15

(b)

Expert Solution

Answer to Problem 5SP

The acceleration of the system is $2.67 m/s2$ .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Given info: The masses of the blocks are $2 kg$ and $6 kg$ .

Write the expression for the acceleration of the horizontal acceleration of the block.

$a=Fnetm$

Here,

$m$ is the total mass of the system

$a$ is the horizontal acceleration of the system

Total mass of the system is $2 kg+4 kg=6 kg$ .

Substitute $16 N$ for $Fnet$ and $6 kg$ for $m$ in the above equation to get $a$ .

$a=16 N6 kg=2.67 m/s2$

Conclusion:

Thus, the acceleration of the system is $2.67 m/s2$ .

Answer 20

(c)

To determine

The force acting on the $2 kg$ mass by the connecting string.

(c)

Expert Solution

Answer to Problem 5SP

The force acting on the $2 kg$ mass by the connecting string is $15.6 N$ .

Explanation of Solution

Given info: The horizontal force acting on $2 kg$ mass along the left direction is $6 N$ .

Let $m1$ be the mass of $2 kg$ block and $m2$ be the mass of $4 kg$ block.

Write the expression for the net force on $2 kg$ mass.

$Fnet1=m1a$

Here,

$m1$ is the mass of $2 kg$ block

$Fnet1$ is the net horizontal force on the mass $2 kg$

Substitute $2 kg$ for $m1$ and $2.67 m/s2$ for $a$ in the above equation to get $Fnet1$ .

$Fnet1=(2 kg )(2.67 m/s2)=5.34 N$

This net force is acting along the direction of acceleration. That is, along the right direction.

Write the expression for the net force on the $2 kg$ mass.

$Fnet1=Ftension−Fleft$

Here,

$Fleft$ is the horizontal force on the left direction of mass $2 kg$

$Ftension$ is the tension on the string

The negative sign indicate that the force of tension is along the right direction whereas the $Fleft$ is along the left direction.

Substitute $6 N$ for $Fleft$ and $5.34 N$ for $Fnet1$ in the above equation to get $Ftension$ .

$5.34 N=Ftension−6 NFtension=11.34N$

Therefore, the force acting on the $2 kg$ by the string is equal to $11.34N$ .

Conclusion:

Thus, the force acting on the $3 kg$ mass by the connecting string is $11.34N$ .

Answer 21

(c)

To determine

The force acting on the $2 kg$ mass by the connecting string.

Answer 22

(c)

Expert Solution

Answer to Problem 5SP

The force acting on the $2 kg$ mass by the connecting string is $15.6 N$ .

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Given info: The horizontal force acting on $2 kg$ mass along the left direction is $6 N$ .

Let $m1$ be the mass of $2 kg$ block and $m2$ be the mass of $4 kg$ block.

Write the expression for the net force on $2 kg$ mass.

$Fnet1=m1a$

Here,

$m1$ is the mass of $2 kg$ block

$Fnet1$ is the net horizontal force on the mass $2 kg$

Substitute $2 kg$ for $m1$ and $2.67 m/s2$ for $a$ in the above equation to get $Fnet1$ .

$Fnet1=(2 kg )(2.67 m/s2)=5.34 N$

This net force is acting along the direction of acceleration. That is, along the right direction.

Write the expression for the net force on the $2 kg$ mass.

$Fnet1=Ftension−Fleft$

Here,

$Fleft$ is the horizontal force on the left direction of mass $2 kg$

$Ftension$ is the tension on the string

The negative sign indicate that the force of tension is along the right direction whereas the $Fleft$ is along the left direction.

Substitute $6 N$ for $Fleft$ and $5.34 N$ for $Fnet1$ in the above equation to get $Ftension$ .

$5.34 N=Ftension−6 NFtension=11.34N$

Therefore, the force acting on the $2 kg$ by the string is equal to $11.34N$ .

Conclusion:

Thus, the force acting on the $3 kg$ mass by the connecting string is $11.34N$ .

Answer 27

(d)

To determine

The net force and acceleration of $4 kg$ block.

(d)

Expert Solution

Answer to Problem 5SP

The net force on the $4 kg$ block is $10.67 N$ and net acceleration of the block is $2.67 m/s2$ .

Explanation of Solution

Given info: The horizontal force acting on $4 kg$ mass along the right direction is $30 N$ and that along the left direction is $8 N$

Write the expression for the net force on the $4 kg$ mass.

$Fnet2=Fright−(Ftension+Fleft)$

Here,

$Fleft$ is the horizontal force on the left direction of mass $4 kg$

$Fright$ is the horizontal force on the right direction of mass $4 kg$

$Ftension$ is the tension on the string

Substitute $30 N$ for $Fright$ , $11.34 N$ for $Ftension$ and $8N$ for $Fleft$ in the above equation to get $Fnet2$ .

$Fnet2=30 N−(11.34 N+8N)=10.66 N$

Write the expression for the net acceleration on $4 kg$ mass.

$a=Fnet2m2$

Here,

$m2$ is the $4 kg$ mass

$a$ is the net acceleration

Substitute $4 kg$ for $m2$ and $10.66 N$ for $Fnet2$ in the above equation to get $Fnet2$ .

$a=10.66 N4 kg=2.67m/s2$

This is same as the net acceleration obtained in part $b$ .

Conclusion:

Thus, the net force on the $4 kg$ block is $10.66 N$ and net acceleration of the block is $2.67m/s2$ $3.2 m/s2$ .

Answer 28

(d)

To determine

The net force and acceleration of $4 kg$ block.

Answer 29

(d)

Expert Solution

Answer to Problem 5SP

The net force on the $4 kg$ block is $10.67 N$ and net acceleration of the block is $2.67 m/s2$ .

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

Given info: The horizontal force acting on $4 kg$ mass along the right direction is $30 N$ and that along the left direction is $8 N$

Write the expression for the net force on the $4 kg$ mass.

$Fnet2=Fright−(Ftension+Fleft)$

Here,

$Fleft$ is the horizontal force on the left direction of mass $4 kg$

$Fright$ is the horizontal force on the right direction of mass $4 kg$

$Ftension$ is the tension on the string

Substitute $30 N$ for $Fright$ , $11.34 N$ for $Ftension$ and $8N$ for $Fleft$ in the above equation to get $Fnet2$ .

$Fnet2=30 N−(11.34 N+8N)=10.66 N$

Write the expression for the net acceleration on $4 kg$ mass.

$a=Fnet2m2$

Here,

$m2$ is the $4 kg$ mass

$a$ is the net acceleration

Substitute $4 kg$ for $m2$ and $10.66 N$ for $Fnet2$ in the above equation to get $Fnet2$ .

$a=10.66 N4 kg=2.67m/s2$

This is same as the net acceleration obtained in part $b$ .

Conclusion:

Thus, the net force on the $4 kg$ block is $10.66 N$ and net acceleration of the block is $2.67m/s2$ $3.2 m/s2$ .

Answer 34

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The net force acting on the entire two-block system.

The net force acting on the entire two-block system.

Answer to Problem 5SP

Explanation of Solution

The acceleration of the system.

Answer to Problem 5SP

Explanation of Solution

The force acting on the 2 kg<m:math><m:mrow><m:mn>2</m:mn><m:mtext> </m:mtext><m:mtext>kg</m:mtext></m:mrow></m:math> mass by the connecting string.

Answer to Problem 5SP

Explanation of Solution

The net force and acceleration of 4 kg<m:math><m:mrow><m:mn>4</m:mn><m:mtext> </m:mtext><m:mtext>kg</m:mtext></m:mrow></m:math> block.

Answer to Problem 5SP

Explanation of Solution

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Chapter 4 Solutions

The force acting on the $2 kg$ mass by the connecting string.

The net force and acceleration of $4 kg$ block.