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CALC The potential energy of two atoms in a diatomic molecule is approximated by U(r) = (a/r12) − (b/r6), where r is the spacing between atoms and a and b are positive constants, (a) Find the force F(r) on one atom as a function of r. Draw two graphs: one of U(r) versus r and one of F(r) versus r. (b) Find the equilibrium distance between the two atoms. Is this equilibrium stable? (c) Suppose the distance between the two atoms is equal to the equilibrium distance found in part (b). What minimum energy must be added to the molecule to dissociate it—that is, to separate the two atoms to an infinite distance apart? This is called the dissociation energy of the molecule, (d) For the molecule CO, the equilibrium distance between the carbon and oxygen atoms is 1.13 × 10 −10 m and the dissociation energy is 1.54 × 10−18 J per molecule. Find the values of the constants a and b.
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- A 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) For what values of the angle between two vectors is their scalar product positive? (b) For what values of is their scalar product negative?arrow_forwardIf 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_forward
- A shopper weighs 3.00 kg of apples on a supermarket scale whose spring obeys Hookes law and notes that the spring stretches a distance of 3.00 cm. a. What will the springs extension be if 5.00 kg of oranges are weighed instead? b. What is the total amount of work that the shopper must do to stretch this spring a total distance of 7.00 cm beyond its relaxed position?arrow_forwardA constant force of magnitude 4.75 N is exerted on an object. The forces direction is 60.0 counterclockwise from the positive x axis in the xy plane, and the objects displacement is r=(4.22.1j+1.6k)m. Calculate the work done by this force.arrow_forwardIf the potential energy of a particle is given by U (r, y, 2) = –kry?2³, find the force associated with this potential energy where k is a positive constant. k(2x²yz³i+y²z³j + xy²2²k) kry²2*(i + ĵ + k) k(y²2³i + 2xyz³j + 3xy²2²k) -kry 2*(i +j+ k) 33 3 k(y²2*i+ xyz°j +xy²2² k)arrow_forward
- In physics, the work done on an object is equal to the integral of the force on that object dotted with its displacent. This looks like W = |(F-ds) (W is work, F is force, and ds is the infinitesimally small displacement vector). For a force whose direction is the line of motion, the equation becomes W = /(Fdz). If the force on an object as a function of displacement is F(z) = 3z² + z, what is the work as a function of displacement W(z)?Assume W(0) = 0 and the force is in the direction of the object's motion.arrow_forwardOne model for the potential energy of a two-atom molecule, where the atoms are separated by a distance r, is U(r)=U0[(r0/r)12−(r0/r)4] where r0 = 0.65 nm and U0 = 7.4 eV.Note: 1 eV = 1.6×10−19 J.Some helpful units:[Force] = eV/nm[Energy] = eV[distance] = nm Here is what I am having trouble with: What is the distance between the atoms when the molecule is in stable equilibrium? If the distance between the atoms increases from equilibrium by r1 = 0.2 nm, then what is the force from one atom on the other associated with this potential energy? (Enter your answer as postive if they repel each other, and negative if they attract.)arrow_forwardRecall that the work WW done by a constant force F⃗ F→ at an angle θθ to the displacement Δr⃗ Δr→ is W=F⃗ ⋅Δr⃗ =FΔrcosθW=F→⋅Δr→=FΔrcosθ. The vector magnitudes FF and ΔrΔr are always positive, so the sign of WW is determined entirely by the angle θθ between the force and the displacement. What is the work WnWn done on the box by the normal force? Express your answers in joules to two significant figures What is the work WTWT done by the tension force? Express your answers in joules to two significant figures.arrow_forward
- Needs Complete typed solution with 100 % accuracy.arrow_forwardRecall that the work WWW done by a constant force F⃗ F→F_vec at an angle θθtheta to the displacement Δr⃗ Δr→ is W=F⃗ ⋅Δr⃗ =FΔrcosθW=F→⋅Δr→=FΔrcosθ. The vector magnitudes FFF and ΔrΔr are always positive, so the sign of WWW is determined entirely by the angle θθtheta between the force and the displacement. What is the work WTWT done by the tension force? Express your answers in joules to two significant figures.arrow_forwardA satellite of the Earth has a mass of 100 kg and is at an altitude of 2.00 ✕ 106 m. (a) What is the gravitational potential energy of the satellite-Earth system? J(b) What is the magnitude of the gravitational force exerted by the Earth on the satellite? N(c) What force does the satellite exert on the Earth? Narrow_forward
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