Prove that the maximum stretching of spring from mean position is x = μ s m g k .

Question

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Answer 1

Question

Chapter 12, Problem 60P

(a)

To determine

Prove that the maximum stretching of spring from mean position is $x=μsmgk$ .

(a)

Expert Solution

Answer to Problem 60P

It is proved that the maximum stretching of spring from mean position is $x=μsmgk$ .

Explanation of Solution

Write the relation between force on spring and maximum static frictional force.

$kx=μsn$

Here, $k$ is the spring force, $x$ is the displacement of spring, $μs$ is the coefficient of static friction, and $n$ is the normal reaction force.

Write the expression for $n$ .

$n=mg$

Here, $m$ is the mass of block and $g$ is the gravitational acceleration.

Rewrite the first equation by substituting the above one.

$kx=μsmg$

Conclusion:

Rewrite the expression for $x$ .

$x=μsmgk$

Therefore, it is proved that the maximum stretching of spring from mean position is $x=μsmgk$ .

$x=μsmgk$ .

(b)

To determine

Prove that block oscillates about mean point when spring is stretched by $x=μsmgk$ .

(b)

Expert Solution

Answer to Problem 60P

It is proved that block oscillates about mean point when spring is stretched by $x=μsmgk$ .

Explanation of Solution

The block will be almost at rest at the break point because of the smaller value of velocity. The block starts to oscillate s soon when the force on block becomes $−kx$ and $+μkmg$ . Write the expression for net force on block while sliding.

$Fnet=−kx+μkmg$

Here, $Fnet$ is the net force on block,.

Rewrite the above equation by multiplying and dividing the second term with $k$ .

$Fnet=−kx+kμkmgk=−k(x−μkmgk)$

Express the term $μkmgk$ into a single variable.

$x−μkmgk=xrel$

Here, $xrel$ is the displacement of block from the point $μkmgk$ .

Rewrite the expression for $Fnet$ by substituting the above relation.

$Fnet=−k(xrel)$

Conclusion:

From the above calculation, it is evident that block starts to be in simple harmonic motion about mean position on displacing the spring

Therefore, it is proved that block oscillates about mean point when spring is stretched by $x=μsmgk$ .

(c)

To determine

Plot the variation of position of block with time.

(c)

Expert Solution

Answer to Problem 60P

The plot is

Principles of Physics: A Calculus-Based Text, Chapter 12, Problem 60P , additional homework tip 1

Explanation of Solution

The graph is plotted with time on x-axis and position on y-axis.

Principles of Physics: A Calculus-Based Text, Chapter 12, Problem 60P , additional homework tip 2

The dotted line along x-axis denotes the function $xrel=μkmgk$ . The peak on curve denotes $x=μsmgk$

Conclusion:

Therefore, the graph is plotted

(d)

To determine

Prove that the amplitude of oscillation is $A=(μs−μk)mgk$ .

(d)

Expert Solution

Answer to Problem 60P

It is proved that the amplitude of oscillation is $A=(μs−μk)mgk$ .

Explanation of Solution

Write the expression for $xrel$ from part (b).

$xrel=x−μkmgk$

Rewrite the above expression by substituting $μsmgk$ for $x$ .

$xrel=μsmgk−μkmgk=(μs−μk)mgk$

Amplitude is the maximum displacement from the mean position only. Rewrite the expression as amplitude.

$A=(μs−μk)mgk$

Here, $A$ is the amplitude.

On pulling the block to distance $x=μsmgk$ , it will start to oscillate about $x=μkmgk$ .

Conclusion:

Therefore, it is proved that the amplitude of oscillation is $A=(μs−μk)mgk$ .

(e)

To determine

Prove that the period of oscillation is $T=2(μs−μk)mgvk+πmk$ .

(e)

Expert Solution

Answer to Problem 60P

It is proved that the period of oscillation is $T=2(μs−μk)mgvk+πmk$ .

Explanation of Solution

Write the expression for time taken by block to move with the board

$T1=2Av$

Here, $T1$ is the time period and $v$ is the speed of motion.

Rewrite the expression by substituting $(μs−μk)mgk$ for $A$ .

$T1=2((μs−μk)mgk)v$

Time taken by block to move from mean position to an extreme position.

$T1/2=12(2πmk)=πmk$

Here, $T1/2$ is the time taken to complete the half of SHM (same as that of time needed to spring back the block to original position.

The period of oscillation is the sum of $T1$ and $T1/2$ .

$T=T1+T1/2$

Here, $T$ is the time period of oscillation.

Conclusion:

Rewrite the above equation by substituting the relations for $T1$ and $T1/2$ .

$T=2((μs−μk)mgk)v+πmk=2((μs−μk)mg)kv+πmk$

Therefore, It is proved that the period of oscillation is $T=2(μs−μk)mgvk+πmk$ .

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Answer 2

Question

Chapter 12, Problem 60P

(a)

To determine

Prove that the maximum stretching of spring from mean position is $x=μsmgk$ .

(a)

Expert Solution

Answer to Problem 60P

It is proved that the maximum stretching of spring from mean position is $x=μsmgk$ .

Explanation of Solution

Write the relation between force on spring and maximum static frictional force.

$kx=μsn$

Here, $k$ is the spring force, $x$ is the displacement of spring, $μs$ is the coefficient of static friction, and $n$ is the normal reaction force.

Write the expression for $n$ .

$n=mg$

Here, $m$ is the mass of block and $g$ is the gravitational acceleration.

Rewrite the first equation by substituting the above one.

$kx=μsmg$

Conclusion:

Rewrite the expression for $x$ .

$x=μsmgk$

Therefore, it is proved that the maximum stretching of spring from mean position is $x=μsmgk$ .

$x=μsmgk$ .

(b)

To determine

Prove that block oscillates about mean point when spring is stretched by $x=μsmgk$ .

(b)

Expert Solution

Answer to Problem 60P

It is proved that block oscillates about mean point when spring is stretched by $x=μsmgk$ .

Explanation of Solution

The block will be almost at rest at the break point because of the smaller value of velocity. The block starts to oscillate s soon when the force on block becomes $−kx$ and $+μkmg$ . Write the expression for net force on block while sliding.

$Fnet=−kx+μkmg$

Here, $Fnet$ is the net force on block,.

Rewrite the above equation by multiplying and dividing the second term with $k$ .

$Fnet=−kx+kμkmgk=−k(x−μkmgk)$

Express the term $μkmgk$ into a single variable.

$x−μkmgk=xrel$

Here, $xrel$ is the displacement of block from the point $μkmgk$ .

Rewrite the expression for $Fnet$ by substituting the above relation.

$Fnet=−k(xrel)$

Conclusion:

From the above calculation, it is evident that block starts to be in simple harmonic motion about mean position on displacing the spring

Therefore, it is proved that block oscillates about mean point when spring is stretched by $x=μsmgk$ .

(c)

To determine

Plot the variation of position of block with time.

(c)

Expert Solution

Answer to Problem 60P

The plot is

Principles of Physics: A Calculus-Based Text, Chapter 12, Problem 60P , additional homework tip 1

Explanation of Solution

The graph is plotted with time on x-axis and position on y-axis.

Principles of Physics: A Calculus-Based Text, Chapter 12, Problem 60P , additional homework tip 2

The dotted line along x-axis denotes the function $xrel=μkmgk$ . The peak on curve denotes $x=μsmgk$

Conclusion:

Therefore, the graph is plotted

(d)

To determine

Prove that the amplitude of oscillation is $A=(μs−μk)mgk$ .

(d)

Expert Solution

Answer to Problem 60P

It is proved that the amplitude of oscillation is $A=(μs−μk)mgk$ .

Explanation of Solution

Write the expression for $xrel$ from part (b).

$xrel=x−μkmgk$

Rewrite the above expression by substituting $μsmgk$ for $x$ .

$xrel=μsmgk−μkmgk=(μs−μk)mgk$

Amplitude is the maximum displacement from the mean position only. Rewrite the expression as amplitude.

$A=(μs−μk)mgk$

Here, $A$ is the amplitude.

On pulling the block to distance $x=μsmgk$ , it will start to oscillate about $x=μkmgk$ .

Conclusion:

Therefore, it is proved that the amplitude of oscillation is $A=(μs−μk)mgk$ .

(e)

To determine

Prove that the period of oscillation is $T=2(μs−μk)mgvk+πmk$ .

(e)

Expert Solution

Answer to Problem 60P

It is proved that the period of oscillation is $T=2(μs−μk)mgvk+πmk$ .

Explanation of Solution

Write the expression for time taken by block to move with the board

$T1=2Av$

Here, $T1$ is the time period and $v$ is the speed of motion.

Rewrite the expression by substituting $(μs−μk)mgk$ for $A$ .

$T1=2((μs−μk)mgk)v$

Time taken by block to move from mean position to an extreme position.

$T1/2=12(2πmk)=πmk$

Here, $T1/2$ is the time taken to complete the half of SHM (same as that of time needed to spring back the block to original position.

The period of oscillation is the sum of $T1$ and $T1/2$ .

$T=T1+T1/2$

Here, $T$ is the time period of oscillation.

Conclusion:

Rewrite the above equation by substituting the relations for $T1$ and $T1/2$ .

$T=2((μs−μk)mgk)v+πmk=2((μs−μk)mg)kv+πmk$

Therefore, It is proved that the period of oscillation is $T=2(μs−μk)mgvk+πmk$ .

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Answer 3

Question

Chapter 12, Problem 60P

(a)

To determine

Prove that the maximum stretching of spring from mean position is $x=μsmgk$ .

(a)

Expert Solution

Answer to Problem 60P

It is proved that the maximum stretching of spring from mean position is $x=μsmgk$ .

Explanation of Solution

Write the relation between force on spring and maximum static frictional force.

$kx=μsn$

Here, $k$ is the spring force, $x$ is the displacement of spring, $μs$ is the coefficient of static friction, and $n$ is the normal reaction force.

Write the expression for $n$ .

$n=mg$

Here, $m$ is the mass of block and $g$ is the gravitational acceleration.

Rewrite the first equation by substituting the above one.

$kx=μsmg$

Conclusion:

Rewrite the expression for $x$ .

$x=μsmgk$

Therefore, it is proved that the maximum stretching of spring from mean position is $x=μsmgk$ .

$x=μsmgk$ .

(b)

To determine

Prove that block oscillates about mean point when spring is stretched by $x=μsmgk$ .

(b)

Expert Solution

Answer to Problem 60P

It is proved that block oscillates about mean point when spring is stretched by $x=μsmgk$ .

Explanation of Solution

The block will be almost at rest at the break point because of the smaller value of velocity. The block starts to oscillate s soon when the force on block becomes $−kx$ and $+μkmg$ . Write the expression for net force on block while sliding.

$Fnet=−kx+μkmg$

Here, $Fnet$ is the net force on block,.

Rewrite the above equation by multiplying and dividing the second term with $k$ .

$Fnet=−kx+kμkmgk=−k(x−μkmgk)$

Express the term $μkmgk$ into a single variable.

$x−μkmgk=xrel$

Here, $xrel$ is the displacement of block from the point $μkmgk$ .

Rewrite the expression for $Fnet$ by substituting the above relation.

$Fnet=−k(xrel)$

Conclusion:

From the above calculation, it is evident that block starts to be in simple harmonic motion about mean position on displacing the spring

Therefore, it is proved that block oscillates about mean point when spring is stretched by $x=μsmgk$ .

(c)

To determine

Plot the variation of position of block with time.

(c)

Expert Solution

Answer to Problem 60P

The plot is

Principles of Physics: A Calculus-Based Text, Chapter 12, Problem 60P , additional homework tip 1

Explanation of Solution

The graph is plotted with time on x-axis and position on y-axis.

Principles of Physics: A Calculus-Based Text, Chapter 12, Problem 60P , additional homework tip 2

The dotted line along x-axis denotes the function $xrel=μkmgk$ . The peak on curve denotes $x=μsmgk$

Conclusion:

Therefore, the graph is plotted

(d)

To determine

Prove that the amplitude of oscillation is $A=(μs−μk)mgk$ .

(d)

Expert Solution

Answer to Problem 60P

It is proved that the amplitude of oscillation is $A=(μs−μk)mgk$ .

Explanation of Solution

Write the expression for $xrel$ from part (b).

$xrel=x−μkmgk$

Rewrite the above expression by substituting $μsmgk$ for $x$ .

$xrel=μsmgk−μkmgk=(μs−μk)mgk$

Amplitude is the maximum displacement from the mean position only. Rewrite the expression as amplitude.

$A=(μs−μk)mgk$

Here, $A$ is the amplitude.

On pulling the block to distance $x=μsmgk$ , it will start to oscillate about $x=μkmgk$ .

Conclusion:

Therefore, it is proved that the amplitude of oscillation is $A=(μs−μk)mgk$ .

(e)

To determine

Prove that the period of oscillation is $T=2(μs−μk)mgvk+πmk$ .

(e)

Expert Solution

Answer to Problem 60P

It is proved that the period of oscillation is $T=2(μs−μk)mgvk+πmk$ .

Explanation of Solution

Write the expression for time taken by block to move with the board

$T1=2Av$

Here, $T1$ is the time period and $v$ is the speed of motion.

Rewrite the expression by substituting $(μs−μk)mgk$ for $A$ .

$T1=2((μs−μk)mgk)v$

Time taken by block to move from mean position to an extreme position.

$T1/2=12(2πmk)=πmk$

Here, $T1/2$ is the time taken to complete the half of SHM (same as that of time needed to spring back the block to original position.

The period of oscillation is the sum of $T1$ and $T1/2$ .

$T=T1+T1/2$

Here, $T$ is the time period of oscillation.

Conclusion:

Rewrite the above equation by substituting the relations for $T1$ and $T1/2$ .

$T=2((μs−μk)mgk)v+πmk=2((μs−μk)mg)kv+πmk$

Therefore, It is proved that the period of oscillation is $T=2(μs−μk)mgvk+πmk$ .

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Answer 4

Question

Chapter 12, Problem 60P

(a)

To determine

Prove that the maximum stretching of spring from mean position is $x=μsmgk$ .

(a)

Expert Solution

Answer to Problem 60P

It is proved that the maximum stretching of spring from mean position is $x=μsmgk$ .

Explanation of Solution

Write the relation between force on spring and maximum static frictional force.

$kx=μsn$

Here, $k$ is the spring force, $x$ is the displacement of spring, $μs$ is the coefficient of static friction, and $n$ is the normal reaction force.

Write the expression for $n$ .

$n=mg$

Here, $m$ is the mass of block and $g$ is the gravitational acceleration.

Rewrite the first equation by substituting the above one.

$kx=μsmg$

Conclusion:

Rewrite the expression for $x$ .

$x=μsmgk$

Therefore, it is proved that the maximum stretching of spring from mean position is $x=μsmgk$ .

$x=μsmgk$ .

(b)

To determine

Prove that block oscillates about mean point when spring is stretched by $x=μsmgk$ .

(b)

Expert Solution

Answer to Problem 60P

It is proved that block oscillates about mean point when spring is stretched by $x=μsmgk$ .

Explanation of Solution

The block will be almost at rest at the break point because of the smaller value of velocity. The block starts to oscillate s soon when the force on block becomes $−kx$ and $+μkmg$ . Write the expression for net force on block while sliding.

$Fnet=−kx+μkmg$

Here, $Fnet$ is the net force on block,.

Rewrite the above equation by multiplying and dividing the second term with $k$ .

$Fnet=−kx+kμkmgk=−k(x−μkmgk)$

Express the term $μkmgk$ into a single variable.

$x−μkmgk=xrel$

Here, $xrel$ is the displacement of block from the point $μkmgk$ .

Rewrite the expression for $Fnet$ by substituting the above relation.

$Fnet=−k(xrel)$

Conclusion:

From the above calculation, it is evident that block starts to be in simple harmonic motion about mean position on displacing the spring

Therefore, it is proved that block oscillates about mean point when spring is stretched by $x=μsmgk$ .

(c)

To determine

Plot the variation of position of block with time.

(c)

Expert Solution

Answer to Problem 60P

The plot is

Principles of Physics: A Calculus-Based Text, Chapter 12, Problem 60P , additional homework tip 1

Explanation of Solution

The graph is plotted with time on x-axis and position on y-axis.

Principles of Physics: A Calculus-Based Text, Chapter 12, Problem 60P , additional homework tip 2

The dotted line along x-axis denotes the function $xrel=μkmgk$ . The peak on curve denotes $x=μsmgk$

Conclusion:

Therefore, the graph is plotted

(d)

To determine

Prove that the amplitude of oscillation is $A=(μs−μk)mgk$ .

(d)

Expert Solution

Answer to Problem 60P

It is proved that the amplitude of oscillation is $A=(μs−μk)mgk$ .

Explanation of Solution

Write the expression for $xrel$ from part (b).

$xrel=x−μkmgk$

Rewrite the above expression by substituting $μsmgk$ for $x$ .

$xrel=μsmgk−μkmgk=(μs−μk)mgk$

Amplitude is the maximum displacement from the mean position only. Rewrite the expression as amplitude.

$A=(μs−μk)mgk$

Here, $A$ is the amplitude.

On pulling the block to distance $x=μsmgk$ , it will start to oscillate about $x=μkmgk$ .

Conclusion:

Therefore, it is proved that the amplitude of oscillation is $A=(μs−μk)mgk$ .

(e)

To determine

Prove that the period of oscillation is $T=2(μs−μk)mgvk+πmk$ .

(e)

Expert Solution

Answer to Problem 60P

It is proved that the period of oscillation is $T=2(μs−μk)mgvk+πmk$ .

Explanation of Solution

Write the expression for time taken by block to move with the board

$T1=2Av$

Here, $T1$ is the time period and $v$ is the speed of motion.

Rewrite the expression by substituting $(μs−μk)mgk$ for $A$ .

$T1=2((μs−μk)mgk)v$

Time taken by block to move from mean position to an extreme position.

$T1/2=12(2πmk)=πmk$

Here, $T1/2$ is the time taken to complete the half of SHM (same as that of time needed to spring back the block to original position.

The period of oscillation is the sum of $T1$ and $T1/2$ .

$T=T1+T1/2$

Here, $T$ is the time period of oscillation.

Conclusion:

Rewrite the above equation by substituting the relations for $T1$ and $T1/2$ .

$T=2((μs−μk)mgk)v+πmk=2((μs−μk)mg)kv+πmk$

Therefore, It is proved that the period of oscillation is $T=2(μs−μk)mgvk+πmk$ .

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Answer 5

Chapter 12, Problem 60P

(a)

To determine

Prove that the maximum stretching of spring from mean position is $x=μsmgk$ .

(a)

Expert Solution

Answer to Problem 60P

It is proved that the maximum stretching of spring from mean position is $x=μsmgk$ .

Explanation of Solution

Write the relation between force on spring and maximum static frictional force.

$kx=μsn$

Here, $k$ is the spring force, $x$ is the displacement of spring, $μs$ is the coefficient of static friction, and $n$ is the normal reaction force.

Write the expression for $n$ .

$n=mg$

Here, $m$ is the mass of block and $g$ is the gravitational acceleration.

Rewrite the first equation by substituting the above one.

$kx=μsmg$

Conclusion:

Rewrite the expression for $x$ .

$x=μsmgk$

Therefore, it is proved that the maximum stretching of spring from mean position is $x=μsmgk$ .

$x=μsmgk$ .

(b)

To determine

Prove that block oscillates about mean point when spring is stretched by $x=μsmgk$ .

(b)

Expert Solution

Answer to Problem 60P

It is proved that block oscillates about mean point when spring is stretched by $x=μsmgk$ .

Explanation of Solution

The block will be almost at rest at the break point because of the smaller value of velocity. The block starts to oscillate s soon when the force on block becomes $−kx$ and $+μkmg$ . Write the expression for net force on block while sliding.

$Fnet=−kx+μkmg$

Here, $Fnet$ is the net force on block,.

Rewrite the above equation by multiplying and dividing the second term with $k$ .

$Fnet=−kx+kμkmgk=−k(x−μkmgk)$

Express the term $μkmgk$ into a single variable.

$x−μkmgk=xrel$

Here, $xrel$ is the displacement of block from the point $μkmgk$ .

Rewrite the expression for $Fnet$ by substituting the above relation.

$Fnet=−k(xrel)$

Conclusion:

From the above calculation, it is evident that block starts to be in simple harmonic motion about mean position on displacing the spring

Therefore, it is proved that block oscillates about mean point when spring is stretched by $x=μsmgk$ .

(c)

To determine

Plot the variation of position of block with time.

(c)

Expert Solution

Answer to Problem 60P

The plot is

Principles of Physics: A Calculus-Based Text, Chapter 12, Problem 60P , additional homework tip 1

Explanation of Solution

The graph is plotted with time on x-axis and position on y-axis.

Principles of Physics: A Calculus-Based Text, Chapter 12, Problem 60P , additional homework tip 2

The dotted line along x-axis denotes the function $xrel=μkmgk$ . The peak on curve denotes $x=μsmgk$

Conclusion:

Therefore, the graph is plotted

(d)

To determine

Prove that the amplitude of oscillation is $A=(μs−μk)mgk$ .

(d)

Expert Solution

Answer to Problem 60P

It is proved that the amplitude of oscillation is $A=(μs−μk)mgk$ .

Explanation of Solution

Write the expression for $xrel$ from part (b).

$xrel=x−μkmgk$

Rewrite the above expression by substituting $μsmgk$ for $x$ .

$xrel=μsmgk−μkmgk=(μs−μk)mgk$

Amplitude is the maximum displacement from the mean position only. Rewrite the expression as amplitude.

$A=(μs−μk)mgk$

Here, $A$ is the amplitude.

On pulling the block to distance $x=μsmgk$ , it will start to oscillate about $x=μkmgk$ .

Conclusion:

Therefore, it is proved that the amplitude of oscillation is $A=(μs−μk)mgk$ .

(e)

To determine

Prove that the period of oscillation is $T=2(μs−μk)mgvk+πmk$ .

(e)

Expert Solution

Answer to Problem 60P

It is proved that the period of oscillation is $T=2(μs−μk)mgvk+πmk$ .

Explanation of Solution

Write the expression for time taken by block to move with the board

$T1=2Av$

Here, $T1$ is the time period and $v$ is the speed of motion.

Rewrite the expression by substituting $(μs−μk)mgk$ for $A$ .

$T1=2((μs−μk)mgk)v$

Time taken by block to move from mean position to an extreme position.

$T1/2=12(2πmk)=πmk$

Here, $T1/2$ is the time taken to complete the half of SHM (same as that of time needed to spring back the block to original position.

The period of oscillation is the sum of $T1$ and $T1/2$ .

$T=T1+T1/2$

Here, $T$ is the time period of oscillation.

Conclusion:

Rewrite the above equation by substituting the relations for $T1$ and $T1/2$ .

$T=2((μs−μk)mgk)v+πmk=2((μs−μk)mg)kv+πmk$

Therefore, It is proved that the period of oscillation is $T=2(μs−μk)mgvk+πmk$ .

Answer 6

Chapter 12, Problem 60P

(a)

To determine

Prove that the maximum stretching of spring from mean position is $x=μsmgk$ .

(a)

Expert Solution

Answer to Problem 60P

It is proved that the maximum stretching of spring from mean position is $x=μsmgk$ .

Explanation of Solution

Write the relation between force on spring and maximum static frictional force.

$kx=μsn$

Here, $k$ is the spring force, $x$ is the displacement of spring, $μs$ is the coefficient of static friction, and $n$ is the normal reaction force.

Write the expression for $n$ .

$n=mg$

Here, $m$ is the mass of block and $g$ is the gravitational acceleration.

Rewrite the first equation by substituting the above one.

$kx=μsmg$

Conclusion:

Rewrite the expression for $x$ .

$x=μsmgk$

Therefore, it is proved that the maximum stretching of spring from mean position is $x=μsmgk$ .

$x=μsmgk$ .

Answer 7

Chapter 12, Problem 60P

(a)

To determine

Prove that the maximum stretching of spring from mean position is $x=μsmgk$ .

Answer 8

(a)

Expert Solution

Answer to Problem 60P

It is proved that the maximum stretching of spring from mean position is $x=μsmgk$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Write the relation between force on spring and maximum static frictional force.

$kx=μsn$

Here, $k$ is the spring force, $x$ is the displacement of spring, $μs$ is the coefficient of static friction, and $n$ is the normal reaction force.

Write the expression for $n$ .

$n=mg$

Here, $m$ is the mass of block and $g$ is the gravitational acceleration.

Rewrite the first equation by substituting the above one.

$kx=μsmg$

Conclusion:

Rewrite the expression for $x$ .

$x=μsmgk$

Therefore, it is proved that the maximum stretching of spring from mean position is $x=μsmgk$ .

$x=μsmgk$ .

Answer 13

(b)

To determine

Prove that block oscillates about mean point when spring is stretched by $x=μsmgk$ .

(b)

Expert Solution

Answer to Problem 60P

It is proved that block oscillates about mean point when spring is stretched by $x=μsmgk$ .

Explanation of Solution

The block will be almost at rest at the break point because of the smaller value of velocity. The block starts to oscillate s soon when the force on block becomes $−kx$ and $+μkmg$ . Write the expression for net force on block while sliding.

$Fnet=−kx+μkmg$

Here, $Fnet$ is the net force on block,.

Rewrite the above equation by multiplying and dividing the second term with $k$ .

$Fnet=−kx+kμkmgk=−k(x−μkmgk)$

Express the term $μkmgk$ into a single variable.

$x−μkmgk=xrel$

Here, $xrel$ is the displacement of block from the point $μkmgk$ .

Rewrite the expression for $Fnet$ by substituting the above relation.

$Fnet=−k(xrel)$

Conclusion:

From the above calculation, it is evident that block starts to be in simple harmonic motion about mean position on displacing the spring

Therefore, it is proved that block oscillates about mean point when spring is stretched by $x=μsmgk$ .

Answer 14

(b)

To determine

Prove that block oscillates about mean point when spring is stretched by $x=μsmgk$ .

Answer 15

(b)

Expert Solution

Answer to Problem 60P

It is proved that block oscillates about mean point when spring is stretched by $x=μsmgk$ .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

The block will be almost at rest at the break point because of the smaller value of velocity. The block starts to oscillate s soon when the force on block becomes $−kx$ and $+μkmg$ . Write the expression for net force on block while sliding.

$Fnet=−kx+μkmg$

Here, $Fnet$ is the net force on block,.

Rewrite the above equation by multiplying and dividing the second term with $k$ .

$Fnet=−kx+kμkmgk=−k(x−μkmgk)$

Express the term $μkmgk$ into a single variable.

$x−μkmgk=xrel$

Here, $xrel$ is the displacement of block from the point $μkmgk$ .

Rewrite the expression for $Fnet$ by substituting the above relation.

$Fnet=−k(xrel)$

Conclusion:

From the above calculation, it is evident that block starts to be in simple harmonic motion about mean position on displacing the spring

Therefore, it is proved that block oscillates about mean point when spring is stretched by $x=μsmgk$ .

Answer 20

(c)

To determine

Plot the variation of position of block with time.

(c)

Expert Solution

Answer to Problem 60P

The plot is

Principles of Physics: A Calculus-Based Text, Chapter 12, Problem 60P , additional homework tip 1

Explanation of Solution

The graph is plotted with time on x-axis and position on y-axis.

Principles of Physics: A Calculus-Based Text, Chapter 12, Problem 60P , additional homework tip 2

The dotted line along x-axis denotes the function $xrel=μkmgk$ . The peak on curve denotes $x=μsmgk$

Conclusion:

Therefore, the graph is plotted

Answer 21

(c)

To determine

Plot the variation of position of block with time.

Answer 22

(c)

Expert Solution

Answer to Problem 60P

The plot is

Principles of Physics: A Calculus-Based Text, Chapter 12, Problem 60P , additional homework tip 1

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

The graph is plotted with time on x-axis and position on y-axis.

Principles of Physics: A Calculus-Based Text, Chapter 12, Problem 60P , additional homework tip 2

The dotted line along x-axis denotes the function $xrel=μkmgk$ . The peak on curve denotes $x=μsmgk$

Conclusion:

Therefore, the graph is plotted

Answer 27

(d)

To determine

Prove that the amplitude of oscillation is $A=(μs−μk)mgk$ .

(d)

Expert Solution

Answer to Problem 60P

It is proved that the amplitude of oscillation is $A=(μs−μk)mgk$ .

Explanation of Solution

Write the expression for $xrel$ from part (b).

$xrel=x−μkmgk$

Rewrite the above expression by substituting $μsmgk$ for $x$ .

$xrel=μsmgk−μkmgk=(μs−μk)mgk$

Amplitude is the maximum displacement from the mean position only. Rewrite the expression as amplitude.

$A=(μs−μk)mgk$

Here, $A$ is the amplitude.

On pulling the block to distance $x=μsmgk$ , it will start to oscillate about $x=μkmgk$ .

Conclusion:

Therefore, it is proved that the amplitude of oscillation is $A=(μs−μk)mgk$ .

Answer 28

(d)

To determine

Prove that the amplitude of oscillation is $A=(μs−μk)mgk$ .

Answer 29

(d)

Expert Solution

Answer to Problem 60P

It is proved that the amplitude of oscillation is $A=(μs−μk)mgk$ .

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

Write the expression for $xrel$ from part (b).

$xrel=x−μkmgk$

Rewrite the above expression by substituting $μsmgk$ for $x$ .

$xrel=μsmgk−μkmgk=(μs−μk)mgk$

Amplitude is the maximum displacement from the mean position only. Rewrite the expression as amplitude.

$A=(μs−μk)mgk$

Here, $A$ is the amplitude.

On pulling the block to distance $x=μsmgk$ , it will start to oscillate about $x=μkmgk$ .

Conclusion:

Therefore, it is proved that the amplitude of oscillation is $A=(μs−μk)mgk$ .

Answer 34

(e)

To determine

Prove that the period of oscillation is $T=2(μs−μk)mgvk+πmk$ .

(e)

Expert Solution

Answer to Problem 60P

It is proved that the period of oscillation is $T=2(μs−μk)mgvk+πmk$ .

Explanation of Solution

Write the expression for time taken by block to move with the board

$T1=2Av$

Here, $T1$ is the time period and $v$ is the speed of motion.

Rewrite the expression by substituting $(μs−μk)mgk$ for $A$ .

$T1=2((μs−μk)mgk)v$

Time taken by block to move from mean position to an extreme position.

$T1/2=12(2πmk)=πmk$

Here, $T1/2$ is the time taken to complete the half of SHM (same as that of time needed to spring back the block to original position.

The period of oscillation is the sum of $T1$ and $T1/2$ .

$T=T1+T1/2$

Here, $T$ is the time period of oscillation.

Conclusion:

Rewrite the above equation by substituting the relations for $T1$ and $T1/2$ .

$T=2((μs−μk)mgk)v+πmk=2((μs−μk)mg)kv+πmk$

Therefore, It is proved that the period of oscillation is $T=2(μs−μk)mgvk+πmk$ .

Answer 35

(e)

To determine

Prove that the period of oscillation is $T=2(μs−μk)mgvk+πmk$ .

Answer 36

(e)

Expert Solution

Answer to Problem 60P

It is proved that the period of oscillation is $T=2(μs−μk)mgvk+πmk$ .

Answer 37

(e)

Expert Solution

Answer 38

(e)

Expert Solution

Answer 39

Expert Solution

Answer 40

Explanation of Solution

Write the expression for time taken by block to move with the board

$T1=2Av$

Here, $T1$ is the time period and $v$ is the speed of motion.

Rewrite the expression by substituting $(μs−μk)mgk$ for $A$ .

$T1=2((μs−μk)mgk)v$

Time taken by block to move from mean position to an extreme position.

$T1/2=12(2πmk)=πmk$

Here, $T1/2$ is the time taken to complete the half of SHM (same as that of time needed to spring back the block to original position.

The period of oscillation is the sum of $T1$ and $T1/2$ .

$T=T1+T1/2$

Here, $T$ is the time period of oscillation.

Conclusion:

Rewrite the above equation by substituting the relations for $T1$ and $T1/2$ .

$T=2((μs−μk)mgk)v+πmk=2((μs−μk)mg)kv+πmk$

Therefore, It is proved that the period of oscillation is $T=2(μs−μk)mgvk+πmk$ .

Answer 41

Ch. 12.1 - A block on the end of a spring is pulled to...Ch. 12.2 - Consider a graphical representation (Fig. 12.3) of...Ch. 12.2 - Figure 12.4 shows two curves representing...Ch. 12.2 - An object of mass m is hung from a spring and set...Ch. 12.4 - A grandfather clock depends on the period of a...Ch. 12.5 - Prob. 12.6QQ Ch. 12 - Which of the following statements is not true...Ch. 12 - Prob. 2OQ Ch. 12 - Prob. 3OQ Ch. 12 - Prob. 4OQ

Prove that the maximum stretching of spring from mean position is x = μ s m g k .

Concept explainers

Videos

Prove that the maximum stretching of spring from mean position is $x=μsmgk$ .

Answer to Problem 60P

Explanation of Solution

Prove that block oscillates about mean point when spring is stretched by $x=μsmgk$ .

Answer to Problem 60P

Explanation of Solution

Plot the variation of position of block with time.

Answer to Problem 60P

Explanation of Solution

Prove that the amplitude of oscillation is $A=(μs−μk)mgk$ .

Answer to Problem 60P

Explanation of Solution

Prove that the period of oscillation is $T=2(μs−μk)mgvk+πmk$ .

Answer to Problem 60P

Explanation of Solution

Want to see more full solutions like this?

Chapter 12 Solutions

Prove that the maximum stretching of spring from mean position is x = μ s m g k .

Concept explainers

Videos

Prove that the maximum stretching of spring from mean position is x=μsmgk<m:math><m:mrow><m:mi>x</m:mi><m:mo>=</m:mo><m:mfrac><m:mrow><m:msub><m:mi>μ</m:mi><m:mi>s</m:mi></m:msub><m:mi>m</m:mi><m:mi>g</m:mi></m:mrow><m:mi>k</m:mi></m:mfrac></m:mrow></m:math>.

Answer to Problem 60P

Explanation of Solution

Prove that block oscillates about mean point when spring is stretched by x=μsmgk<m:math><m:mrow><m:mi>x</m:mi><m:mo>=</m:mo><m:mfrac><m:mrow><m:msub><m:mi>μ</m:mi><m:mi>s</m:mi></m:msub><m:mi>m</m:mi><m:mi>g</m:mi></m:mrow><m:mi>k</m:mi></m:mfrac></m:mrow></m:math>.

Answer to Problem 60P

Explanation of Solution

Plot the variation of position of block with time.

Answer to Problem 60P

Explanation of Solution

Answer to Problem 60P

Explanation of Solution

Answer to Problem 60P

Explanation of Solution

Want to see more full solutions like this?

Chapter 12 Solutions

Prove that the maximum stretching of spring from mean position is $x=μsmgk$ .

Prove that block oscillates about mean point when spring is stretched by $x=μsmgk$ .