Physics for Scientists and Engineers
Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
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Chapter 7, Problem 31P

(a)

To determine

The points where object is in equilibrium.

(a)

Expert Solution
Check Mark

Explanation of Solution

Given:

The potential energy function is U=3x2x3 for x3.0m .

The potential energy function is U=0 for x3.0m .

Formula used:

  U(x) is the potential energy of the particle and Fx is negative of derivative of potential energy function.

Write the expression for force in terms of potential energy.

  Fx=dUdx ........(1)

Here, Fx is the force acting on the object, dU is change of potential energy and dx is change in position.

Calculation:

For x3.0m :

Substitute 3x2x3 for U in equation (1).

  Fx=ddx(3x2x3)=3x(2x)

Calculate the equilibrium points.

  Fx=03x(2x)=0x=0andx=2.0m

For x3.0m :

Substitute 0 for U in equation (1).

  Fx=ddx(0)=0

Object would be in neutral equilibrium for x3.0m .

Conclusion:

Thus, object is in equilibrium at x=0,x=2.0m and x3.0m .

(b)

To determine

The graph for U versus x .

(b)

Expert Solution
Check Mark

Explanation of Solution

Given:

The potential energy function is U=3x2x3 for x3.0m .

The potential energy function is U=0 for x3.0m .

Formula used:

The energy contained in an object due to its position when compared to other objects is called potential energy.

Write the expression of potential energy for x3.0m .

  U=3x2x3 ........(2)

Here, U is potential energy, x is the position of object.

Write the expression of potential energy for x3.0m .

  U=0

Calculation:

Substitute 1.0m for x in equation (2).

  U=(3.0J/ m 2)(1.0m)2(1.0J/ m 2)(1.0m)3=4.0J

Substitute 0m for x in equation (2).

  U=(3.0J/ m 2)(0m)2(1.0J/ m 2)(0m)3=0J

Substitute 1.0m for x in equation (2).

  U=(3.0J/ m 2)(1.0m)2(1.0J/ m 2)(1.0m)3=2.0J

Substitute 2.0m for x in equation (2).

  U=(3.0J/ m 2)(2.0m)2(1.0J/ m 2)(2.0m)3=4.0J

Substitute 3.0m for x in equation (2).

  U=(3.0J/ m 2)(3.0m)2(1.0J/ m 2)(3.0m)3=0J

The graph for U versus x is plotted below.

  Physics for Scientists and Engineers, Chapter 7, Problem 31P

Conclusion:

Thus, the graph for U versus x is plotted above.

(c)

To determine

The stable and unstable points of equilibrium.

(c)

Expert Solution
Check Mark

Explanation of Solution

Given:

The potential energy function is U=3x2x3 for x3.0m .

The potential energy function is U=0 for x3.0m .

Formula used:

  U(x) is the potential energy of the particle and Fx is negative of derivative of potential energy function.

Write the expression for force in terms of potential energy.

  Fx=dUdx ........(1)

Here, Fx is the force acting on the object, dU is change of potential energy and dx is change in position.

Calculation:

For x3.0m :

Substitute 3x2x3 for U in equation (1).

  Fx=ddx(3x2x3)=3x(2x)

Calculate the equilibrium points.

  Fx=03x(2x)=0x=0andx=2.0m

For x3.0m :

Substitute 0 for U in equation (1).

  Fx=ddx(0)=0

Object would be in neutral equilibrium for x3.0m .

Calculate the potential energy at all equilibrium points.

For x=0 :

  U=(3.0J/ m 2)(0m)2(1.0J/ m 2)(0m)3=0J

For x=2.0m :

  U=(3.0J/ m 2)(2.0m)2(1.0J/ m 2)(2.0m)3=4.0J

Conclusion:

The value of potential energy is maximum at x=2.0m and minimum at x=0 . Thus, the stable equilibrium point is x=0 and the unstable equilibrium point is x=2.0m.

(d)

To determine

The speed of object at x=2.0m .

(d)

Expert Solution
Check Mark

Explanation of Solution

Given:

The potential energy function is U=3x2x3 for x3.0m .

The potential energy function is U=0 for x3.0m .

The mass of object is 4.0kg .

The total energy of object is 12J at x=2.0m .

Formula used:

Write the expression for potential energy.

  U=3x2x3 ........(2)

Here, x is the position of object and U is the potential energy.

Write the expression for change in kinetic energy.

  K=12mv2

Here, m is mass of object, v is velocity of object and K is the kinetic energy.

Write the expression of total energy.

  E=U+K ........(3)

Here, E is the total energy.

Substitute 12mv2 for K in equation (3) and rearrange in terms of v .

  v=2( EU)m ........(4)

Calculation:

Substitute 2.0m for x in equation (2).

  U=(3.0J/ m 2)(2.0m)2(1.0J/ m 2)(2.0m)3=4.0J

Substitute 12J for E , 4.0J for U and 4.0kg for m in equation (4).

  v= 2( 12J4.0J ) 4.0kg=2.0m/s

Conclusion:

The speed of object at x=2.0m is 2.0m/s .

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Chapter 7 Solutions

Physics for Scientists and Engineers

Ch. 7 - Prob. 11PCh. 7 - Prob. 12PCh. 7 - Prob. 13PCh. 7 - Prob. 14PCh. 7 - Prob. 15PCh. 7 - Prob. 16PCh. 7 - Prob. 17PCh. 7 - Prob. 18PCh. 7 - Prob. 19PCh. 7 - Prob. 20PCh. 7 - Prob. 21PCh. 7 - Prob. 22PCh. 7 - Prob. 23PCh. 7 - Prob. 24PCh. 7 - Prob. 25PCh. 7 - Prob. 26PCh. 7 - Prob. 27PCh. 7 - Prob. 28PCh. 7 - Prob. 29PCh. 7 - Prob. 30PCh. 7 - Prob. 31PCh. 7 - Prob. 32PCh. 7 - Prob. 33PCh. 7 - Prob. 34PCh. 7 - Prob. 35PCh. 7 - Prob. 36PCh. 7 - Prob. 37PCh. 7 - Prob. 38PCh. 7 - Prob. 39PCh. 7 - Prob. 40PCh. 7 - Prob. 41PCh. 7 - Prob. 42PCh. 7 - Prob. 43PCh. 7 - Prob. 44PCh. 7 - Prob. 45PCh. 7 - Prob. 46PCh. 7 - Prob. 47PCh. 7 - Prob. 48PCh. 7 - Prob. 49PCh. 7 - Prob. 50PCh. 7 - Prob. 51PCh. 7 - Prob. 52PCh. 7 - Prob. 53PCh. 7 - Prob. 54PCh. 7 - Prob. 55PCh. 7 - Prob. 56PCh. 7 - Prob. 57PCh. 7 - Prob. 58PCh. 7 - Prob. 59PCh. 7 - Prob. 60PCh. 7 - Prob. 61PCh. 7 - Prob. 62PCh. 7 - Prob. 63PCh. 7 - Prob. 64PCh. 7 - Prob. 65PCh. 7 - Prob. 66PCh. 7 - Prob. 67PCh. 7 - Prob. 68PCh. 7 - Prob. 69PCh. 7 - Prob. 70PCh. 7 - Prob. 71PCh. 7 - Prob. 72PCh. 7 - Prob. 73PCh. 7 - Prob. 74PCh. 7 - Prob. 75PCh. 7 - Prob. 76PCh. 7 - Prob. 77PCh. 7 - Prob. 78PCh. 7 - Prob. 79PCh. 7 - Prob. 80PCh. 7 - Prob. 81PCh. 7 - Prob. 82PCh. 7 - Prob. 83PCh. 7 - Prob. 84PCh. 7 - Prob. 85PCh. 7 - Prob. 86PCh. 7 - Prob. 87PCh. 7 - Prob. 88PCh. 7 - Prob. 89PCh. 7 - Prob. 90PCh. 7 - Prob. 91PCh. 7 - Prob. 92PCh. 7 - Prob. 93PCh. 7 - Prob. 94PCh. 7 - Prob. 95PCh. 7 - Prob. 96PCh. 7 - Prob. 97PCh. 7 - Prob. 98PCh. 7 - Prob. 99PCh. 7 - Prob. 100PCh. 7 - Prob. 101PCh. 7 - Prob. 102PCh. 7 - Prob. 103PCh. 7 - Prob. 104PCh. 7 - Prob. 105PCh. 7 - Prob. 106P
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