The condition that determines l 1 and to show that the net force on the mass is F = − k x .

Question

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Answer 1

Question

Chapter 5, Problem 5.1P

(a)

To determine

The condition that determines l1 and to show that the net force on the mass is F=−kx.

(a)

Expert Solution

Answer to Problem 5.1P

The net force acting on the spring is Fnet=−kx

Explanation of Solution

The stretched length of the spring l1 when the mass is attached to it is determined by the force constant k of the spring and the mass m that is hung to it. These are the conditions that determent stretched length l1.

The net force acting on the spring when the mass is attached to it and it is in equilibrium position is zero because the downward pull of gravity on the mass is counter equally by the spring force in opposite direction.

mg+[−k(l1−l0)]=0 (I)

Here, m is the mass, g is the acceleration due to gravity, and k is the spring constant.

Now, when the spring is pulled further down by a distance x beyond its new equilibrium position, then net force on the mass is,

Fnet=−k(l1−l0)+mg+(−kx)=0+(−kx)=−kx

Substitute 0 for −k(l1−l0)+mg in above equation.

Fnet=0+(−kx)=−kx

Conclusion:

Therefore, the net force acting on the spring is Fnet=−kx

(b)

To determine

To prove the net potential energy has the form U(x)=12kx2+const.

(b)

Expert Solution

Answer to Problem 5.1P

The net potential energy has the form U(x)=12kx2+const.

Explanation of Solution

The stretching force is conservative as the net stretching force depends only on the distance the spring is stretched, that is x.

Write the expression for the potential energy.

U(x)=−∫0xFnet(x)dx

Substitute −kx for Fnet(x) in the above equation and simplify.

U(x)=−∫0xFnet(x)dx=−∫0x(−kx)dx=k∫0xxdx=k(x22)0x+Const=12kx2+Const

Conclusion:

Therefore, the net potential energy has the form U(x)=12kx2+const.

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Answer 2

Question

Chapter 5, Problem 5.1P

(a)

To determine

The condition that determines l1 and to show that the net force on the mass is F=−kx.

(a)

Expert Solution

Answer to Problem 5.1P

The net force acting on the spring is Fnet=−kx

Explanation of Solution

The stretched length of the spring l1 when the mass is attached to it is determined by the force constant k of the spring and the mass m that is hung to it. These are the conditions that determent stretched length l1.

The net force acting on the spring when the mass is attached to it and it is in equilibrium position is zero because the downward pull of gravity on the mass is counter equally by the spring force in opposite direction.

mg+[−k(l1−l0)]=0 (I)

Here, m is the mass, g is the acceleration due to gravity, and k is the spring constant.

Now, when the spring is pulled further down by a distance x beyond its new equilibrium position, then net force on the mass is,

Fnet=−k(l1−l0)+mg+(−kx)=0+(−kx)=−kx

Substitute 0 for −k(l1−l0)+mg in above equation.

Fnet=0+(−kx)=−kx

Conclusion:

Therefore, the net force acting on the spring is Fnet=−kx

(b)

To determine

To prove the net potential energy has the form U(x)=12kx2+const.

(b)

Expert Solution

Answer to Problem 5.1P

The net potential energy has the form U(x)=12kx2+const.

Explanation of Solution

The stretching force is conservative as the net stretching force depends only on the distance the spring is stretched, that is x.

Write the expression for the potential energy.

U(x)=−∫0xFnet(x)dx

Substitute −kx for Fnet(x) in the above equation and simplify.

U(x)=−∫0xFnet(x)dx=−∫0x(−kx)dx=k∫0xxdx=k(x22)0x+Const=12kx2+Const

Conclusion:

Therefore, the net potential energy has the form U(x)=12kx2+const.

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Answer 3

Question

Chapter 5, Problem 5.1P

(a)

To determine

The condition that determines l1 and to show that the net force on the mass is F=−kx.

(a)

Expert Solution

Answer to Problem 5.1P

The net force acting on the spring is Fnet=−kx

Explanation of Solution

The stretched length of the spring l1 when the mass is attached to it is determined by the force constant k of the spring and the mass m that is hung to it. These are the conditions that determent stretched length l1.

The net force acting on the spring when the mass is attached to it and it is in equilibrium position is zero because the downward pull of gravity on the mass is counter equally by the spring force in opposite direction.

mg+[−k(l1−l0)]=0 (I)

Here, m is the mass, g is the acceleration due to gravity, and k is the spring constant.

Now, when the spring is pulled further down by a distance x beyond its new equilibrium position, then net force on the mass is,

Fnet=−k(l1−l0)+mg+(−kx)=0+(−kx)=−kx

Substitute 0 for −k(l1−l0)+mg in above equation.

Fnet=0+(−kx)=−kx

Conclusion:

Therefore, the net force acting on the spring is Fnet=−kx

(b)

To determine

To prove the net potential energy has the form U(x)=12kx2+const.

(b)

Expert Solution

Answer to Problem 5.1P

The net potential energy has the form U(x)=12kx2+const.

Explanation of Solution

The stretching force is conservative as the net stretching force depends only on the distance the spring is stretched, that is x.

Write the expression for the potential energy.

U(x)=−∫0xFnet(x)dx

Substitute −kx for Fnet(x) in the above equation and simplify.

U(x)=−∫0xFnet(x)dx=−∫0x(−kx)dx=k∫0xxdx=k(x22)0x+Const=12kx2+Const

Conclusion:

Therefore, the net potential energy has the form U(x)=12kx2+const.

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Answer 4

Question

Chapter 5, Problem 5.1P

(a)

To determine

The condition that determines l1 and to show that the net force on the mass is F=−kx.

(a)

Expert Solution

Answer to Problem 5.1P

The net force acting on the spring is Fnet=−kx

Explanation of Solution

The stretched length of the spring l1 when the mass is attached to it is determined by the force constant k of the spring and the mass m that is hung to it. These are the conditions that determent stretched length l1.

The net force acting on the spring when the mass is attached to it and it is in equilibrium position is zero because the downward pull of gravity on the mass is counter equally by the spring force in opposite direction.

mg+[−k(l1−l0)]=0 (I)

Here, m is the mass, g is the acceleration due to gravity, and k is the spring constant.

Now, when the spring is pulled further down by a distance x beyond its new equilibrium position, then net force on the mass is,

Fnet=−k(l1−l0)+mg+(−kx)=0+(−kx)=−kx

Substitute 0 for −k(l1−l0)+mg in above equation.

Fnet=0+(−kx)=−kx

Conclusion:

Therefore, the net force acting on the spring is Fnet=−kx

(b)

To determine

To prove the net potential energy has the form U(x)=12kx2+const.

(b)

Expert Solution

Answer to Problem 5.1P

The net potential energy has the form U(x)=12kx2+const.

Explanation of Solution

The stretching force is conservative as the net stretching force depends only on the distance the spring is stretched, that is x.

Write the expression for the potential energy.

U(x)=−∫0xFnet(x)dx

Substitute −kx for Fnet(x) in the above equation and simplify.

U(x)=−∫0xFnet(x)dx=−∫0x(−kx)dx=k∫0xxdx=k(x22)0x+Const=12kx2+Const

Conclusion:

Therefore, the net potential energy has the form U(x)=12kx2+const.

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Answer 5

Chapter 5, Problem 5.1P

(a)

To determine

The condition that determines l1 and to show that the net force on the mass is F=−kx.

(a)

Expert Solution

Answer to Problem 5.1P

The net force acting on the spring is Fnet=−kx

Explanation of Solution

The stretched length of the spring l1 when the mass is attached to it is determined by the force constant k of the spring and the mass m that is hung to it. These are the conditions that determent stretched length l1.

The net force acting on the spring when the mass is attached to it and it is in equilibrium position is zero because the downward pull of gravity on the mass is counter equally by the spring force in opposite direction.

mg+[−k(l1−l0)]=0 (I)

Here, m is the mass, g is the acceleration due to gravity, and k is the spring constant.

Now, when the spring is pulled further down by a distance x beyond its new equilibrium position, then net force on the mass is,

Fnet=−k(l1−l0)+mg+(−kx)=0+(−kx)=−kx

Substitute 0 for −k(l1−l0)+mg in above equation.

Fnet=0+(−kx)=−kx

Conclusion:

Therefore, the net force acting on the spring is Fnet=−kx

(b)

To determine

To prove the net potential energy has the form U(x)=12kx2+const.

(b)

Expert Solution

Answer to Problem 5.1P

The net potential energy has the form U(x)=12kx2+const.

Explanation of Solution

The stretching force is conservative as the net stretching force depends only on the distance the spring is stretched, that is x.

Write the expression for the potential energy.

U(x)=−∫0xFnet(x)dx

Substitute −kx for Fnet(x) in the above equation and simplify.

U(x)=−∫0xFnet(x)dx=−∫0x(−kx)dx=k∫0xxdx=k(x22)0x+Const=12kx2+Const

Conclusion:

Therefore, the net potential energy has the form U(x)=12kx2+const.

Answer 6

Chapter 5, Problem 5.1P

(a)

To determine

The condition that determines l1 and to show that the net force on the mass is F=−kx.

(a)

Expert Solution

Answer to Problem 5.1P

The net force acting on the spring is Fnet=−kx

Explanation of Solution

The stretched length of the spring l1 when the mass is attached to it is determined by the force constant k of the spring and the mass m that is hung to it. These are the conditions that determent stretched length l1.

The net force acting on the spring when the mass is attached to it and it is in equilibrium position is zero because the downward pull of gravity on the mass is counter equally by the spring force in opposite direction.

mg+[−k(l1−l0)]=0 (I)

Here, m is the mass, g is the acceleration due to gravity, and k is the spring constant.

Now, when the spring is pulled further down by a distance x beyond its new equilibrium position, then net force on the mass is,

Fnet=−k(l1−l0)+mg+(−kx)=0+(−kx)=−kx

Substitute 0 for −k(l1−l0)+mg in above equation.

Fnet=0+(−kx)=−kx

Conclusion:

Therefore, the net force acting on the spring is Fnet=−kx

Answer 7

Chapter 5, Problem 5.1P

(a)

To determine

The condition that determines l1 and to show that the net force on the mass is F=−kx.

Answer 8

(a)

Expert Solution

Answer to Problem 5.1P

The net force acting on the spring is Fnet=−kx

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

The stretched length of the spring l1 when the mass is attached to it is determined by the force constant k of the spring and the mass m that is hung to it. These are the conditions that determent stretched length l1.

The net force acting on the spring when the mass is attached to it and it is in equilibrium position is zero because the downward pull of gravity on the mass is counter equally by the spring force in opposite direction.

mg+[−k(l1−l0)]=0 (I)

Here, m is the mass, g is the acceleration due to gravity, and k is the spring constant.

Now, when the spring is pulled further down by a distance x beyond its new equilibrium position, then net force on the mass is,

Fnet=−k(l1−l0)+mg+(−kx)=0+(−kx)=−kx

Substitute 0 for −k(l1−l0)+mg in above equation.

Fnet=0+(−kx)=−kx

Conclusion:

Therefore, the net force acting on the spring is Fnet=−kx

Answer 13

(b)

To determine

To prove the net potential energy has the form U(x)=12kx2+const.

(b)

Expert Solution

Answer to Problem 5.1P

The net potential energy has the form U(x)=12kx2+const.

Explanation of Solution

The stretching force is conservative as the net stretching force depends only on the distance the spring is stretched, that is x.

Write the expression for the potential energy.

U(x)=−∫0xFnet(x)dx

Substitute −kx for Fnet(x) in the above equation and simplify.

U(x)=−∫0xFnet(x)dx=−∫0x(−kx)dx=k∫0xxdx=k(x22)0x+Const=12kx2+Const

Conclusion:

Therefore, the net potential energy has the form U(x)=12kx2+const.

Answer 14

(b)

To determine

To prove the net potential energy has the form U(x)=12kx2+const.

Answer 15

(b)

Expert Solution

Answer to Problem 5.1P

The net potential energy has the form U(x)=12kx2+const.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

The stretching force is conservative as the net stretching force depends only on the distance the spring is stretched, that is x.

Write the expression for the potential energy.

U(x)=−∫0xFnet(x)dx

Substitute −kx for Fnet(x) in the above equation and simplify.

U(x)=−∫0xFnet(x)dx=−∫0x(−kx)dx=k∫0xxdx=k(x22)0x+Const=12kx2+Const

Conclusion:

Therefore, the net potential energy has the form U(x)=12kx2+const.

Answer 20

Ch. 5 - Prob. 5.1P Ch. 5 - Prob. 5.2P Ch. 5 - Prob. 5.3P Ch. 5 - Prob. 5.4P Ch. 5 - Prob. 5.5P Ch. 5 - Prob. 5.6P Ch. 5 - Prob. 5.7P Ch. 5 - Prob. 5.8P Ch. 5 - Prob. 5.9P Ch. 5 - Prob. 5.10P

The condition that determines l 1 and to show that the net force on the mass is F = − k x .

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The condition that determines l1 and to show that the net force on the mass is F=−kx.

Answer to Problem 5.1P

Explanation of Solution

To prove the net potential energy has the form U(x)=12kx2+const.

Answer to Problem 5.1P

Explanation of Solution

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Chapter 5 Solutions