Physics for Scientists and Engineers: Foundations and Connections
1st Edition
ISBN: 9781133939146
Author: Katz, Debora M.
Publisher: Cengage Learning
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Chapter 9.6, Problem 9.3CE
(a)
To determine
Find the force which causes the man to accelerate.
(b)
To determine
Find out whether this force does any work on the man or not.
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Chapter 9 Solutions
Physics for Scientists and Engineers: Foundations and Connections
Ch. 9.4 - In the three cases shown in Figure 9.11, a force...Ch. 9.6 - Prob. 9.2CECh. 9.6 - Prob. 9.3CECh. 9.7 - Prob. 9.4CECh. 9.7 - Prob. 9.5CECh. 9.9 - Prob. 9.6CECh. 9 - Pick an isolated system for the following...Ch. 9 - Prob. 2PQCh. 9 - Prob. 3PQCh. 9 - Prob. 4PQ
Ch. 9 - Prob. 5PQCh. 9 - Prob. 6PQCh. 9 - Prob. 7PQCh. 9 - A 537-kg trailer is hitched to a truck. Find the...Ch. 9 - Prob. 9PQCh. 9 - A helicopter rescues a trapped person of mass m =...Ch. 9 - Prob. 11PQCh. 9 - An object is subject to a force F=(512i134j) N...Ch. 9 - Prob. 13PQCh. 9 - Prob. 14PQCh. 9 - Prob. 15PQCh. 9 - Prob. 16PQCh. 9 - Prob. 17PQCh. 9 - Prob. 18PQCh. 9 - Prob. 19PQCh. 9 - Prob. 20PQCh. 9 - Prob. 21PQCh. 9 - Prob. 22PQCh. 9 - A constant force of magnitude 4.75 N is exerted on...Ch. 9 - In three cases, a force acts on a particle, and...Ch. 9 - An object of mass m = 5.8 kg moves under the...Ch. 9 - A nonconstant force is exerted on a particle as it...Ch. 9 - Prob. 27PQCh. 9 - Prob. 28PQCh. 9 - Prob. 29PQCh. 9 - A particle moves in the xy plane (Fig. P9.30) from...Ch. 9 - A small object is attached to two springs of the...Ch. 9 - Prob. 32PQCh. 9 - Prob. 33PQCh. 9 - Prob. 34PQCh. 9 - Prob. 35PQCh. 9 - Prob. 36PQCh. 9 - Prob. 37PQCh. 9 - Prob. 38PQCh. 9 - A shopper weighs 3.00 kg of apples on a...Ch. 9 - Prob. 40PQCh. 9 - Prob. 41PQCh. 9 - Prob. 42PQCh. 9 - Prob. 43PQCh. 9 - Prob. 44PQCh. 9 - Prob. 45PQCh. 9 - Prob. 46PQCh. 9 - Prob. 47PQCh. 9 - Prob. 48PQCh. 9 - Prob. 49PQCh. 9 - A small 0.65-kg box is launched from rest by a...Ch. 9 - A small 0.65-kg box is launched from rest by a...Ch. 9 - A horizontal spring with force constant k = 625...Ch. 9 - A box of mass m = 2.00 kg is dropped from rest...Ch. 9 - Prob. 54PQCh. 9 - Return to Example 9.9 and use the result to find...Ch. 9 - Prob. 56PQCh. 9 - Crall and Whipple design a loop-the-loop track for...Ch. 9 - Prob. 58PQCh. 9 - Calculate the force required to pull a stuffed toy...Ch. 9 - Prob. 60PQCh. 9 - Prob. 61PQCh. 9 - Prob. 62PQCh. 9 - An elevator motor moves a car with six people...Ch. 9 - Prob. 64PQCh. 9 - Figure P9.65A shows a crate attached to a rope...Ch. 9 - Prob. 66PQCh. 9 - Prob. 67PQCh. 9 - Prob. 68PQCh. 9 - Prob. 69PQCh. 9 - Prob. 70PQCh. 9 - Prob. 71PQCh. 9 - Estimate the power required for a boxer to jump...Ch. 9 - Prob. 73PQCh. 9 - Prob. 74PQCh. 9 - Prob. 75PQCh. 9 - Prob. 76PQCh. 9 - Prob. 77PQCh. 9 - Prob. 78PQCh. 9 - Prob. 79PQCh. 9 - A block of mass m = 0.250 kg is pressed against a...Ch. 9 - On a movie set, an alien spacecraft is to be...Ch. 9 - Prob. 82PQCh. 9 - A spring-loaded toy gun is aimed vertically and...Ch. 9 - Prob. 84PQCh. 9 - The motion of a box of mass m = 2.00 kg along the...Ch. 9 - Prob. 86PQCh. 9 - Prob. 87PQCh. 9 - Prob. 88PQ
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- Suppose the ski patrol lowers a rescue sled and victim, having a total mass of 90.0 kg, down a 60.0° slope at constant speed, as shown in Figure 7.37. The coefficient of friction between the sled and the snow is 0.100. (a) How much work is done by friction as the sled moves 30.0 m along the hill? (b) How much work is done by the rope on the sled in this distance? (c) What is the work done by the gravitational force on the sled? (d) What is the total work done?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_forward(a) Suppose a constant force acts on an object. The force does not vary with time or with the position or the velocity of the object. Start with the general definition for work done by a force W=ifFdr and show that the force is conservative, (b) As a special case, suppose the force F =(3i + 4j)N acts on a particle that moves from O to in Figure P7.43. Calculate the work done by F on the particle as it moves along each one of the three paths shown in the figure and show that the work done along the three paths identical.arrow_forward
- . In the annual Empire State Building race, contestants run up 1,575 steps to a height of 1,050 ft. In 2003, Australian Paul Crake completed the race in a record time of 9 min and 33 S, Mr., Crake weighed 143 lb (65 kg) , (a) How much work did Mr., Crake do in reaching the top of the building? (b) What was his average power output (in ft-lb/s and in hp)?arrow_forward(a) How long will it take an 850-kg car with a useful power output of 40.0 hp (1hp=746W) to reach a speed of 15.0 m/s, neglecting friction? (b) How long will this acceleration take if the car also climbs a 3.00-m-high hill in the process?arrow_forwardAs a simple pendulum swings back and forth, the forces acting on the suspended object are the force of gravity, the tension in the supporting cord, and air resistance, (a) Which of these forces, if any, does no work on the pendulum? (b) Which of these forces does negative work at all times during the pendulums motion? (c) Describe the work done by the force of gravity while the pendulum is swinging.arrow_forward
- A shopper pushes a grocery cart 20.0 m at constant speed on level ground, against a 35.0 N frictional force. He pushes in a direction 25.0° below the horizontal. (a) What is the work done on the cart by friction? (b) What is the work done on the cart by the gravitational force? (c) What is the work done on the cart by the shopper? (d) Find the force the shopper exerts, using energy considerations. (e) What is the total work done on the cart?arrow_forwardA nonconstant force is exerted on a particle as it moves in the positive direction along the x axis. Figure P9.26 shows a graph of this force Fx versus the particles position x. Find the work done by this force on the particle as the particle moves as follows. a. From xi = 0 to xf = 10.0 m b. From xi = 10.0 to xf = 20.0 m c. From xi = 0 to xf = 20.0 m FIGURE P9.26 Problems 26 and 27.arrow_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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