10. An object moves in the x-y plane subject to a force F(x.y) = y i + (x + y) j How much work is done by this force if the object moves in a straight line from the point (0,1) to the point (1,2)? (Hint: you must find the equation of the line.) 11. Mars has mass of 6.42 × 1023 kg and a radius of 3,380 km. a) What is the acceleration of gravity at the surface of Mars? b) How long would it take an object to fall 10.0 meters from rest on Mars? c) Phobos, a moon of Mars, orbits Mars at a distance of 6,000 km from the surface of Mars. What is its orbital speed

10. An object moves in the x-y plane subject to a force F(x.y) = y i + (x + y) j How much work is done by this force if the object moves in a straight line from the point (0,1) to the point (1,2)? (Hint: you must find the equation of the line.) 11. Mars has mass of 6.42 × 1023 kg and a radius of 3,380 km. a) What is the acceleration of gravity at the surface of Mars? b) How long would it take an object to fall 10.0 meters from rest on Mars? c) Phobos, a moon of Mars, orbits Mars at a distance of 6,000 km from the surface of Mars. What is its orbital speed

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 42P: A 4.00-kg particle moves from the origin to position ©, having coordinates x = 5.00 m and y = 5.00 m...

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Question

10 and 11 please

10. An object moves in the x-y plane subject to a force F(x.y) = y i + (x + y) j
How much work is done by this force if the object moves in a straight line from the point (0,1) to the point (1,2)?
(Hint: you must find the equation of the line.)
11. Mars has mass of 6.42 × 1023 kg and a radius of 3,380 km.
a) What is the acceleration of gravity at the surface of Mars?
b) How long would it take an object to fall 10.0 meters from rest on Mars?
c) Phobos, a moon of Mars, orbits Mars at a distance of 6,000 km from the surface of Mars. What is its orbital speed

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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?
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A 4.00-kg particle moves from the origin to position , having coordinates x = 5.00 m and y = 5.00 m (Fig. P7.31). One force on the particle is the gravitational force acting in the negative y direction. Using Equation 7.3, 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 P7.31
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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