Problem1. A block slides on a rough horizontal surface from point A to point B. A force (magnitude P = 2.0 N) acts on the block between A and B, as shown. Points A and B are 1.5 m apart. (a) How much work is done on the block by the force P? (b) If the kinetic energies of the block at A and B are 6.0 J and 4.0 J, respectively, how much work is done on the block by the net force as the block moves from A to B? (c) What is the work done by the friction force? 40° B

Problem1. A block slides on a rough horizontal surface from point A to point B. A force (magnitude P = 2.0 N) acts on the block between A and B, as shown. Points A and B are 1.5 m apart. (a) How much work is done on the block by the force P? (b) If the kinetic energies of the block at A and B are 6.0 J and 4.0 J, respectively, how much work is done on the block by the net force as the block moves from A to B? (c) What is the work done by the friction force? 40° B

Name: Problem1. A block slides on a rough horizontal surface from point A t
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Description: Problem1. A block slides on a rough horizontal surface from point A to point B. A force (magnitude P= 2.0 N) acts on the block between A and B, as shown. Points A and B are 1.5 m apart.(a) How much work is done on the block by the force P?(b) If the

Physics for Scientists and Engineers: Foundations and Connections

1st Edition

ISBN:9781133939146

Author:Katz, Debora M.

Publisher:Katz, Debora M.

Chapter9: Energy In Nonisolated Systems

Section: Chapter Questions

Problem 28PQ

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A particle is subject to a force Fx that varies with position as shown in Figure P7.9. Find the work done by the force on the particle as it moves (a) from x = 0 to x = 5.00 m, (b) from x = 5.00 m to x = 10.0 m, and (c) from x = 10.0 m to x = 15.0 m. (d) What is the total work done by the force over the distance x = 0 to x = 15.0 m?
a shopper in a supermarket pushes a cart with a force of 35 N directed at an angle of 25 below the horizontal. The force is just sufficient to overcome various frictional forces, so the cart moves at constant speed, (a) Find the work done by the shopper as she moves down a 50.0-m length aisle, (b) What is the net work done on the cart? Why? (c) The shopper goes down the next aisle, pushing horizontally and maintaining the same speed as before. If the work done by frictional forces doesnt change, would the shoppers applied force be larger, smaller, or the same? What about the work done on the cart by the shopper?
a shopper in a supermarket pushes a cart with a force of 35 N directed at an angle of 25 below the horizontal. The force is just sufficient to overcome various frictional forces, so the cart moves at constant speed, (a) Find the work done by the shopper as she moves down a 50.0-m length aisle, (b) What is the net work done on the cart? Why? (c) The shopper goes down the next aisle, pushing horizontally and maintaining the same speed as before. If the work done by frictional forces doesnt change, would the shoppers applied force be larger, smaller, or the same? What about the work done on the cart by the shopper?
Physics Review A team of huskies performs 7 440 J of work on a loaded sled of mass 124 kg, drawing it from rest up a 4.60-m high snow-covered rise while the sled loses 1 520 J due to friction, (a) What is the net work done on the sled by the huskies and friction? (b) What is the change in the sleds potential energy? (c) What is the speed of the sled at the top of the rise? (See Section 5.5.)
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A 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.
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Give an example of a situation in which there is a force and a displacement, but the force does no work. Explain why it does no work.
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 P9.30ANS
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?