University Physics Volume 1
18th Edition
ISBN: 9781938168277
Author: William Moebs, Samuel J. Ling, Jeff Sanny
Publisher: OpenStax - Rice University
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Textbook Question
Chapter 8, Problem 27P
The potential energy function for either one of the two atoms in a diatomic molecule is often approximated by
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27.The potential energy function for either one of the two atoms in a diatomic molecule is often
approximated by U (x)=-a/x¹² -b/x6 where x is the distance between the atoms. (a) At what
distance of seperation does the potential energy have a local minimum (not at x = ¥)?(b) What
is the force on an atom at this separation? (c) How does the force vary with the separation
distance?
Solution
a.
-2a
b
1/6
; b. 0; c. ~x6
3) The equation that describes the potential energy stored in the interaction between two atoms in a diatomic molecule is U = A/x12 – B/x6. Here A and B are appropriate constants and x is the inter-atomic distance. Find the magnitude of the force that one atom exerts on the other.
a) 12A/x13 – 6B/x7
b) –13A/x13 + 7B/x7
c) –11A/x11 + 5B/x5
d) 72A/x12 – 72B/x6
e) A/x13 – B/x7
The potential energy of a nitrogen atom in an ammonia (NH3) molecule varies with position as U(x) = x4 - 2ax2 (a is some constant)
a) Give an expression for the force F on the nitrogen atom and sketch F(x).
b) At what positions will the nitrogen atom be in stable equilibrium?
Chapter 8 Solutions
University Physics Volume 1
Ch. 8 - Check Your understanding In Example 8.1 what are...Ch. 8 - Check Your Understanding What are the values of...Ch. 8 - Check Your Understanding When the length of the...Ch. 8 - Check Your Understanding Suppose the mass in...Ch. 8 - Check Your Understanding A two-dimensional,...Ch. 8 - Check Your Understanding Fend the forces on the...Ch. 8 - Check Your Understanding How high above the bottom...Ch. 8 - Check Your Understanding You probably recall that,...Ch. 8 - Check Your Understanding What potential energy...Ch. 8 - Check Your Understanding Repeat Example 8.10 when...
Ch. 8 - Check Your Understanding Find x(t) for the...Ch. 8 - The kinetic energy of a system must always be...Ch. 8 - The force exerted by a diving board is...Ch. 8 - Describe the gravitational potential energy...Ch. 8 - A couple of soccer balls of equal mass are kiched...Ch. 8 - What is the dominant factor that affects the speed...Ch. 8 - Two people observe a leaf falling from a tree. One...Ch. 8 - What is the physical meaning of a non-conservative...Ch. 8 - A bottle rocket is shot straight up in the air...Ch. 8 - An external force acts on a particle during a trip...Ch. 8 - When a body slides down an inclined plane, does...Ch. 8 - Consider the following scenario. 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What...Ch. 8 - Repeat the preceding problem, but this time,...Ch. 8 - Two bodies are interacting by a conservative force...Ch. 8 - In an amusement park, a car rolls in a track as...Ch. 8 - A 200-g steel ball is tied to a 2.00m “massless”...Ch. 8 - A 300 g hockey puck is shot across an ice-covered...Ch. 8 - A projectile of mass 2 kg is fired with a speed of...Ch. 8 - An artillery shell is fired at a target 200 m...Ch. 8 - How much energy is lost to a dissipative drag...Ch. 8 - A box slides on a frictionless surface with a...
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- If the net work done by external forces on a particle is zero, which of the following statements about the particle must be true? (a) Its velocity is zero. (b) Its velocity is decreased. (c) Its velocity is unchanged. (d) Its speed is unchanged. (e) More information is needed.arrow_forwardWhen a 4.00-kg object is hung vertically on a certain light spring that obeys Hookes law, the spring stretches 2.50 cm. If the 4.00-kg object is removed, (a) how far will the spring stretch if a 1.50-kg block is hung on it? (b) How much work must an external agent do to stretch the same spring 4.00 cm from its unstretched position?arrow_forwardA block of mass m = 2.50 kg is pushed a distance d = 2.20 m along a frictionless, horizontal table by a constant applied force of magnitude F = 16.0 N directed at an angle = 25.0 below the horizontal as shown in Figure P6.3. Determine the work done on the block by (a) the applied force, (b) the normal force exerted by the table, (c) the gravitational force, and (d) the net force on the block. Figure P6.3arrow_forward
- A 4.00-kg particle moves from the origin to position , having coordinates x = 5.00 m and y = 5.00 m (Fig. P7.31). One force on the particle is the gravitational force acting in the negative y direction. Using Equation 7.3, calculate the work done by the gravitational force on the particle as it goes from O to along (a) the purple path, (b) the red path, and (c) the blue path, (d) Your results should all be identical. Why? Figure P7.31arrow_forwardA particle moves in the xy plane (Fig. P9.30) from the origin to a point having coordinates x = 7.00 m and y = 4.00 m under the influence of a force given by F=3y2+x. a. What is the work done on the particle by the force F if it moves along path 1 (shown in red)? b. What is the work done on the particle by the force F if it moves along path 2 (shown in blue)? c. What is the work done on the particle by the force F if it moves along path 3 (shown in green)? d. Is the force F conservative or nonconservative? Explain. FIGURE P9.30 In each case, the work is found using the integral of Fdr along the path (Equation 9.21). W=rtrfFdr=rtrf(Fxdx+Fydy+Fzdz) (a) The work done along path 1, we first need to integrate along dr=dxi from (0,0) to (7,0) and then along dr=dyj from (7,0) to (7,4): W1=x=0;y=0x=7;y=0(3y2i+xj)(dxi)+x=7;y=0x=7;y=4(3y2i+xj)(dyj) Performing the dot products, we get W1=x=0;y=0x=7;y=03y2dx+x=7;y=0x=7;y=4xdy Along the first part of this path, y = 0 therefore the first integral equals zero. For the second integral, x is constant and can be pulled out of the integral, and we can evaluate dy. W1=0+x=7;y=0x=7;y=4xdy=xy|x=7;y=0x=7;y=4=28J (b) The work done along path 2 is along dr=dyj from (0,0) to (0,4) and then along dr=dxi from (0,4) to (7,4): W2=x=0;y=0x=0;y=4(3y2i+xj)(dyj)+x=0;y=4x=7;y=4(3y2i+xj)(dyi) Performing the dot product, we get: W2=x=0;y=0x=0;y=4xdy+x=0;y=4x=7;y=43y2dx Along the first part of this path, x = 0. Therefore, the first integral equals zero. For the second integral, y is constant and can be pulled out of the integral, and we can evaluate dx. W2=0+3y2x|x=0;y=4x=7;y=4=336J (c) To find the work along the third path, we first write the expression for the work integral. W=rtrfFdr=rtrf(Fxdx+Fydy+Fzdz)W=rtrf(3y2dx+xdy)(1) At first glance, this appears quite simple, but we cant integrate xdy=xy like we might have above because the value of x changes as we vary y (i.e., x is a function of y.) [In parts (a) and (b), on a straight horizontal or vertical line, only x or y changes]. One approach is to parameterize both x and y as a function of another variable, say t, and write each integral in terms of only x or y. Constraining dr to be along the desired line, we can relate dx and dy: tan=dydxdy=tandxanddx=dytan(2) Now, use equation (2) in (1) to express each integral in terms of only one variable. W=x=0;y=0x=7;y=43y2dx+x=0;y=0x=7;y=4xdyW=y=0y=43y2dytan+x=0x=7xtandx We can determine the tangent of the angle, which is constant (the angle is the angle of the line with respect to the horizontal). tan=4.007.00=0.570 Insert the value of the tangent and solve the integrals. W=30.570y33|y=0y=4+0.570x22|x=0x=7W=112+14=126J (d) Since the work done is not path-independent, this is non-conservative force. Figure P9.30ANSarrow_forward(a) Can the kinetic energy of a system be negative? (b) Can the gravitational potential energy of a system be negative? Explain.arrow_forward
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