FUNDAMENTALS OF PHYSICS - EXTENDED
12th Edition
ISBN: 9781119773511
Author: Halliday
Publisher: WILEY
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Chapter 7, Problem 63P
63 SSM To push a 25.0 kg crate up a frictionless incline, angled at 25.0° to the horizontal, a worker exerts a force of 209 N parallel to the incline. As the crate slides 1.50 m, how much work is done on the crate by (a) the worker’s applied force, (b) the gravitational force on the crate, and (c) the normal force exerted by the incline on the crate? (d) What is the total work done on the crate?
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Chapter 7 Solutions
FUNDAMENTALS OF PHYSICS - EXTENDED
Ch. 7 - Prob. 1QCh. 7 - Figure 7-16a shows two horizontal forces that act...Ch. 7 - Is positive or negative work done by a constant...Ch. 7 - Spring A is stiffer than spring B kA kB. The...Ch. 7 - In three situations, a single force acts on a...Ch. 7 - Figure 7-23 shows three arrangements of a block...Ch. 7 - SSM A proton mass m = 1.67 1027 kg is being...Ch. 7 - If a Saturn V rocket with an Apollo spacecraft...Ch. 7 - On August 10, 1972, a large meteorite skipped...Ch. 7 - An explosion at ground level leaves a crater with...
Ch. 7 - A father racing his son has half the kinetic...Ch. 7 - A ice block floating in a river is pushed through...Ch. 7 - The only force acting on a 2.0 kg canister that is...Ch. 7 - A coin slides over a frictionless plane and across...Ch. 7 - A 12.0 N force with a fixed orientation does work...Ch. 7 - A luge and its rider, with a total mass of 85 kg,...Ch. 7 - SSM WWW A helicopter lifts a 72 kg astronaut 15 m...Ch. 7 - a In 1975 the roof of Montreals Velodrome, witha...Ch. 7 - 21 SSM A cord is used to vertically lower an...Ch. 7 - A cave rescue team lifts an injured spelunker...Ch. 7 - In Fig. 7-10, we must apply a force of magnitude...Ch. 7 - During spring semester at MIT, residents of the...Ch. 7 - In Fig. 7-10a, a block of mass m lies on a...Ch. 7 - SSM WWW The only force acting on a 2.0 kg body as...Ch. 7 - SSM WWW The force on a particle is directed along...Ch. 7 - A 1.5 kg block is initially at rest on a...Ch. 7 - GO A force F= cx3.00x2iacts on a particle as the...Ch. 7 - A can of sardines is made to move along an xaxis...Ch. 7 - A single force acts on a 3.0 kg particle-like...Ch. 7 - GO Figure 7-41 shows a cord attached to a cart...Ch. 7 - SSM A force of 5.0 N acts on a 15 kg body...Ch. 7 - A skier is pulled by a towrope up a frictionless...Ch. 7 - SSM ILW A 100 kg block is pulled at a constant...Ch. 7 - The loaded cab of an elevator has a mass of 3.0 ...Ch. 7 - A machine carries a 4.0 kg package from an initial...Ch. 7 - A 0.30 kg ladle sliding on a horizontal...Ch. 7 - Prob. 49PCh. 7 - a At a certain instant, a particle-like object is...Ch. 7 - A force F= 3.00 N i 7.00 N j 7.00 N k acts on...Ch. 7 - A funny car accelerates from rest through a...Ch. 7 - SSM A horse pulls a cart with a force of 40 lb at...Ch. 7 - An initially stationary 2.0 kg object accelerates...Ch. 7 - To pull a 50 kg crate across a horizontal...Ch. 7 - A frightened child is restrained by her mother as...Ch. 7 - How much work is done by a force F= 2x N i 3 N j,...Ch. 7 - 63 SSM To push a 25.0 kg crate up a frictionless...Ch. 7 - Boxes are transported from one location to another...Ch. 7 - If a car of mass 1200 kg is moving along a highway...Ch. 7 - An iceboat is at rest on a frictionless frozen...Ch. 7 - If a ski lift raises 100 passengers averaging 660...Ch. 7 - A force F= 4.0 N i cj acts on a particle as the...Ch. 7 - Prob. 71PCh. 7 - Prob. 72PCh. 7 - Prob. 73PCh. 7 - Prob. 74PCh. 7 - Prob. 75PCh. 7 - Prob. 76PCh. 7 - Prob. 77PCh. 7 - Prob. 78P
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- A 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_forwardAs a young man, Tarzan climbed up a vine to reach his tree house. As he got older, he decided to build and use a staircase instead. Since the work of the gravitational force mg is path Independent, what did the King of the Apes gain in using stairs?arrow_forward(a) A force F=(4xi+3yj), where F is in newtons and x and y are in meters, acts on an object as the object moves in the x direction from the origin to x = 5.00 m. Find the work W=Fdr done by the force on the object. (b) What If? Find the work W=Fdr done by the force on the object if it moves from the origin to (5.00 m, 5.00 m) along a straightline path making an angle of 45.0 with the positive x axis. Is the work done by this force dependent on the path taken between the initial and final points?arrow_forward
- A 4.00-kg particle moves from the origin to position ©, having coordinates x = 5.00 m and y = 5.00 m (Fig. P6.42). One force on the particle is the gravitational force acting in the negative y direction. Using Equation 6.3, 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 P6.42 Problems 42 through 45.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_forwardA 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_forward
- 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_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_forward
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