Physics for Scientists and Engineers with Modern Physics, Technology Update
9th Edition
ISBN: 9781305401969
Author: SERWAY, Raymond A.; Jewett, John W.
Publisher: Cengage Learning
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Chapter 13, Problem 35P
A system consists of three particles, each of mass 5.00 g, located at the corners of an equilateral triangle with sides of 30.0 cm. (a) Calculate the gravitational potential energy of the system. (b) Assume the particles are released simultaneously. Describe the subsequent motion of each. Will any collisions take place? Explain.
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Chapter 13 Solutions
Physics for Scientists and Engineers with Modern Physics, Technology Update
Ch. 13.1 - A planet has two moons of equal mass. Moon 1 is in...Ch. 13.2 - Prob. 13.2QQCh. 13.4 - Prob. 13.3QQCh. 13.6 - Prob. 13.4QQCh. 13 - Prob. 1OQCh. 13 - Prob. 2OQCh. 13 - Prob. 3OQCh. 13 - Prob. 4OQCh. 13 - Prob. 5OQCh. 13 - Prob. 6OQ
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- A system consists of three particles, each of mass 5.00 g, located at the corners of an equilateral triangle with sides of 30.0 cm. (a) Calculate the gravitational potential energy of the system. (b) Assume the particles are released simultaneously. Describe the subsequent motion of each. Will any collisions take place? Explain.arrow_forward(a) Can the kinetic energy of a system be negative? (b) Can the gravitational potential energy of a system be negative? Explain.arrow_forwardRank the following quantities of energy from largest to the smallest. State if any are equal. (a) the absolute value of the average potential energy of the SunEarth system (b) the average kinetic energy of the Earth in its orbital motion relative to the Sun (c) the absolute value of the total energy of the SunEarth systemarrow_forward
- In each situation shown in Figure P8.12, a ball moves from point A to point B. Use the following data to find the change in the gravitational potential energy in each case. You can assume that the radius of the ball is negligible. a. h = 1.35 m, = 25, and m = 0.65 kg b. R = 33.5 m and m = 756 kg c. R = 33.5 m and m = 756 kg FIGURE P8.12 Problems 12, 13, and 14.arrow_forwardAn inclined plane of angle = 20.0 has a spring of force constant k = 500 N/m fastened securely at the bottom so that the spring is parallel to the surface as shown in Figure P6.61. A block of mass m = 2.50 kg is placed on the plane at a distance d = 0.300 m from the spring. From this position, the block is projected downward toward the spring with speed v = 0.750 m/s. By what distance is the spring compressed when the block momentarily comes to rest?arrow_forwardA small block of mass m = 200 g is released from rest at point along the horizontal diameter on the inside of a frictionless, hemispherical bowl of radius R = 30.0 cm (Fig. P7.45). Calculate (a) the gravitational potential energy of the block-Earth system when the block is at point relative to point . (b) the kinetic energy of the block at point , (c) its speed at point , and (d) its kinetic energy and the potential energy when the block is at point . Figure P7.45 Problems 45 and 46.arrow_forward
- A suspicious physics student watches a stunt performed at an ice show. In the stunt, a performer shoots an arrow into a bale of hay (Fig. P11.24). Another performer rides on the bale of hay like a cowboy. After the arrow enters the bale, the balearrow system slides roughly 5 m along the ice. Estimate the initial speed of the arrow. Is there a trick to this stunt? FIGURE P11.24arrow_forwardIn a laboratory experiment, an electron with a kinetic energy of 50.5 keV is shot toward another electron initially at rest (Fig. P11.50). (1 eV = 1.602 1019 J) The collision is elastic. The initially moving electron is deflected by the collision. a. Is it possible for the initially stationary electron to remain at rest after the collision? Explain. b. The initially moving electron is detected at an angle of 40.0 from its original path. What is the speed of each electron after the collision? FIGURE P11.50arrow_forwardA space probe is fired as a projectile from the Earths surface with an initial speed of 2.00 104 m/s. What will its speed be when it is very far from the Earth? Ignore atmospheric friction and the rotation of the Earth. P11.26 Ki+Ui=Kf+Uf12mvi2+GMEm(1rf1ri)=12mvf212vi2+GME(01RE)=12vf2orvf2=v122GMEREandvf=(v122GMERE)1/2,vf=[(2.00104)21.25108]1/2m/s=1.66104m/sarrow_forward
- Estimate the kinetic energy of the following: a. An ant walking across the kitchen floor b. A baseball thrown by a professional pitcher c. A car on the highway d. A large truck on the highwayarrow_forwardA block is placed on top of a vertical spring, and the spring compresses. Figure P8.24 depicts a moment in time when the spring is compressed by an amount h. a. To calculate the change in the gravitational and elastic potential energies, what must be included in the system? b. Find an expression for the change in the systems potential energy in terms of the parameters shown in Figure P8.24. c. If m = 0.865 kg and k = 125 N/m, find the change in the systems potential energy when the blocks displacement is h = 0.0650 m, relative to its initial position. FIGURE P8.24arrow_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
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