A webassign.net 12. DETAILS SERPSE10 7.4.P.009. MY NOT A particle is subject to a force F, that varies with position as shown in the following figure. F, (N) 3. 1 х (m) 6 8 10 12 14 16 2 4 (a) Find the work done by the force on the particle as it moves from x = 0 to x 3.00 m. (b) Find the work done by the force on the particle as it moves from x = 5.00 m to x = 9.00 m. (c) Find the work done by the force on the particle as it moves from x 11.0 m tox = 15.0 m. %3D (d) What is the total work done by the force over the distance x = 0 to x = 15.0 m? Need Help? Read It Watch It 13. [5/10 Points] DETAILS PREVIOUS ANSWERS SERPSE10 7.4.P.016. MY NOTES esc 名口 888 F3 FS @ $ % & 4 9 Q E R T Y A D F G K V B M ait option command # 00 出 #3 2. 2.

A webassign.net 12. DETAILS SERPSE10 7.4.P.009. MY NOT A particle is subject to a force F, that varies with position as shown in the following figure. F, (N) 3. 1 х (m) 6 8 10 12 14 16 2 4 (a) Find the work done by the force on the particle as it moves from x = 0 to x 3.00 m. (b) Find the work done by the force on the particle as it moves from x = 5.00 m to x = 9.00 m. (c) Find the work done by the force on the particle as it moves from x 11.0 m tox = 15.0 m. %3D (d) What is the total work done by the force over the distance x = 0 to x = 15.0 m? Need Help? Read It Watch It 13. [5/10 Points] DETAILS PREVIOUS ANSWERS SERPSE10 7.4.P.016. MY NOTES esc 名口 888 F3 FS @ $ % & 4 9 Q E R T Y A D F G K V B M ait option command # 00 出 #3 2. 2.

Name: A webassign.net12.DETAILSSERPSE10 7.4.P.009.MY NOTA particle is subje
Uploaded: 2024-03-20T06:57:25.000Z
Channel: Bartleby
Description: A webassign.net12.DETAILSSERPSE10 7.4.P.009.MY NOTA particle is subject to a force F, that varies with position as shown in the following figure.F, (N)3.1х (m)6 8 10 12 14 162 4(a) Find the work done by the force on the particle as it moves from x =

Principles of Physics: A Calculus-Based Text

5th Edition

ISBN:9781133104261

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter6: Energy Of A System

Section: Chapter Questions

Problem 15P

See similar textbooks

Similar questions

(a) What is the average useful power output of a person who does 6.00106J of useful work in 8.00 h? (b) Working at this rate, how long will it take this person to lift 2000 kg of bricks 1.50 m to a platform? (Work done to lift his body can be omitted because it is not considered useful output here.)
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?
As 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?
The force acting on a particle varies as shown in Figure P6.14. Find the work done by the force on the particle as it moves (a) from x = 0 to x = 8.00 m, (b) from x = 8.00 m to x= 10.0 m, and (c) from x = 0 to x = 10.0 m.
A person in good physical condition can put out 100 W of useful power for several hours at a stretch, perhaps by pedaling a mechanism that drives an electric generator. Neglecting any problems of generator efficiency and practical considerations such as resting time: (a) How many people ‘would it take to nm a 4.O0-kW electric clothes dryer? (b) How many people would it take to replace a large electric power plant that generates 800 MW?
(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?
An archer pulls her bowstring back 0.400 m by exerting a force that increases uniformly from zero to 230 N. (a) What is the equivalent spring constant of the bow? (b) How much work does the archer do on the string in drawing the bow?
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?
The force acting on a panicle varies as shown in Figure la P7.14. Find the work done by the force on the particle as it moves (a) from x = 0 to x = 8.00 m. (b) from x = 8.00 m to x = 10.0 m, and (c) from x = 0 to x = 10.0 m.
(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?
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
Suppose the ski patrol lowers a rescue sled and victim, having a total mass of 90.0 kg, down a 60.00 slope at constant speed, as shown below. The coefficient of friction between the sled and the snow is 0.100. (a) How much work is done by fiction 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?