Physics for Scientists and Engineers with Modern Physics
10th Edition
ISBN: 9781337553292
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 7, Problem 15P
A small particle of mass m is pulled to the top of a friction less half-cylinder (of radius R) by a light cord that passes over the top of the cylinder as illustrated in Figure P7.15. (a) Assuming the particle moves at a constant speed, show that F = mg cos θ. Note: If the particle moves at constant speed, the component of its acceleration tangent to the cylinder must be zero at all times. (b) By directly integrating
Figure P7.15
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In a truck-loading station at a post office, a small 0.200 kg package is released from rest at point A on a track that is one-quarter of a circle with radius 1.60 m. The size of the package is much less than 1.60 m, so the package can be treated as a particle. It slides down the track and reaches point B with a speed of 4.80 m/s. From point B, it slides on a level surface a distance of 3.00 m to point C, where it comes to rest. (a) What is the coefficient of kinetic friction on the horizontal surface? (b) How much work is done on the package by friction as it slides down the circular arc from A to B?
Chapter 7 Solutions
Physics for Scientists and Engineers with Modern Physics
Ch. 7.2 - Prob. 7.1QQCh. 7.2 - shows four situations in which a force is applied...Ch. 7.3 - Which of the following statements is true about...Ch. 7.4 - A dart is inserted into a spring-loaded dart gun...Ch. 7.5 - A dart is inserted into a spring-loaded dart gun...Ch. 7.6 - Choose the correct answer. The gravitational...Ch. 7.6 - A ball is connected to a light spring suspended...Ch. 7.8 - What does the slope of a graph of U(x) versus x...Ch. 7 - A shopper in a supermarket pushes a cart with a...Ch. 7 - The record number of boat lifts, including the...
Ch. 7 - In 1990, Walter Arfeuille of Belgium lifted a...Ch. 7 - Spiderman, whose mass is 80.0 kg, is dangling on...Ch. 7 - Prob. 5PCh. 7 - Vector A has a magnitude of 5.00 units, and vector...Ch. 7 - Find the scalar product of the vectors in Figure...Ch. 7 - Using the definition of the scalar product, find...Ch. 7 - A particle is subject to a force Fx that varies...Ch. 7 - In a control system, an accelerometer consists of...Ch. 7 - When a 4.00-kg object is hung vertically on a...Ch. 7 - Prob. 12PCh. 7 - The tray dispenser in your cafeteria has broken...Ch. 7 - Prob. 14PCh. 7 - A small particle of mass m is pulled to the top of...Ch. 7 - The force acting on a particle is Fx = (8x 16),...Ch. 7 - When different loads hang on a spring, the spring...Ch. 7 - Prob. 18PCh. 7 - (a) A force F=(4xi+3yj), where F is in newtons and...Ch. 7 - Review. The graph in Figure P7.20 specifies a...Ch. 7 - A 0.600-kg particle has a speed of 2.00 m/s at...Ch. 7 - A 4.00-kg particle is subject to a net force that...Ch. 7 - A 2 100-kg pile driver is used to drive a steel...Ch. 7 - Review. In an electron microscope, there is an...Ch. 7 - Review. You can think of the workkinetic energy...Ch. 7 - You are lying in your bedroom, resting after doing...Ch. 7 - Review. A 5.75-kg object passes through the origin...Ch. 7 - Prob. 28PCh. 7 - A 0.20-kg stone is held 1.3 m above the top edge...Ch. 7 - A 1 000-kg roller coaster car is initially at the...Ch. 7 - A 4.00-kg particle moves from the origin to...Ch. 7 - (a) Suppose a constant force acts on an object....Ch. 7 - A force acting on a particle moving in the xy...Ch. 7 - Prob. 34PCh. 7 - Prob. 35PCh. 7 - Prob. 36PCh. 7 - Prob. 37PCh. 7 - For the potential energy curve shown in Figure...Ch. 7 - A right circular cone can theoretically be...Ch. 7 - The potential energy function for a system of...Ch. 7 - You have a new internship, where you are helping...Ch. 7 - Prob. 42APCh. 7 - A particle moves along the xaxis from x = 12.8 m...Ch. 7 - Why is the following situation impossible? In a...Ch. 7 - Prob. 45APCh. 7 - Prob. 46APCh. 7 - An inclined plane of angle = 20.0 has a spring of...Ch. 7 - Prob. 48APCh. 7 - Over the Christmas break, you are making some...Ch. 7 - A particle of mass m = 1.18 kg is attached between...
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- A 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_forwardIn 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_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
- A system consists of five particles. How many terms appear in the expression for the total gravitational potential energy of the system? (a) 4 (b) 5 (c) 10 (d) 20 (e) 25arrow_forwardA particle moves along a curved path y(x)=(10m){1+cos[0.1m1]x} , from x=0 to x=10m , subject to a tangential force of variable magnitude F(x)=(10N)sin[(0.1m-1)x] . How much work does the force do? (Hint: Consult a table of integrals or use a numerical integration program.)arrow_forwardA roller-coaster car shown in Figure P7.82 is released from rest from a height h and then moves freely with negligible friction. The roller-coaster track includes a circular loop of radius R in a vertical plane. (a) First suppose the car barely makes it around the loop; at the top of the loop, the riders are upside down and feel weightless. Find the required height h of the release point above the bottom of the loop in terms of R. (b) Now assume the release point is at or above the minimum required height. Show that the normal force on the car at the bottom of the loop exceeds the normal force at the top of the loop by six times the cars weight. The normal force on each rider follows the same rule. Such a large normal force is dangerous and very uncomfortable for the riders. Roller coasters are therefore not built with circular loops in vertical planes. Figure P5.22 (page 149) shows an actual design.arrow_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 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_forwardA 4.0-kg particle moving along the x -axis is acted upon by the force whose functional form appears below. The velocity of the particle at x = 0 is v = 6.0 m/s. Find the particle’s speed at x=(a)2.0m, 2.0 (b)4.0 m. (c) 10.0m, (d) Does the particle turn around at some point and head back toward the origin? (e) Repeat part (d) if v = 2.0 m/s at x = 0.arrow_forward
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