A 4500 kg spacecraft is in a circular orbit 1500 km above the surtace of Mars (my = 6.42 x 1023 kg. Ry= 3.39 x 10 m). Part A How much work must the spacecraft engines perform to move the spacecraft to a circular orbit that is 4500 km above the surface? Express your answer with the appropriate units. W = 1.49 • 100

A 4500 kg spacecraft is in a circular orbit 1500 km above the surtace of Mars (my = 6.42 x 1023 kg. Ry= 3.39 x 10 m). Part A How much work must the spacecraft engines perform to move the spacecraft to a circular orbit that is 4500 km above the surface? Express your answer with the appropriate units. W = 1.49 • 100

Name: E ReviewA 4500 kg spacecraft is in a circular orbit 1500 km above the
Uploaded: 2024-03-20T06:57:16.000Z
Channel: Bartleby
Description: E ReviewA 4500 kg spacecraft is in a circular orbit 1500 km above thesurface of Mars (my = 6.42 x 1023 kg,Ry = 3.39 x 10 m).Part AHow much work must the spacecraft engines perform to move the spacecraft to a circular orbit that is 4500 km above the

Physics for Scientists and Engineers: Foundations and Connections

1st Edition

ISBN:9781133939146

Author:Katz, Debora M.

Publisher:Katz, Debora M.

Chapter8: Conservation Of Energy

Section: Chapter Questions

Problem 26PQ: A block is hung from a vertical spring. The spring stretches (h = 0.0650 m) as shown for a...

See similar textbooks

Similar questions

(a) Can the kinetic energy of a system be negative? (b) Can the gravitational potential energy of a system be negative? Explain.
(a) A child slides down a water slide at an amusement park from an initial height h. The slide can be considered frictionless because of the water flowing down it. Can the equation for conservation of mechanical energy be used on the child? (b) Is the mass of the child a factor in determining his speed at the bottom of the slide? (c) The child drops straight down rather than following the curved ramp of the slide. In which case will he be traveling faster at ground level? (d) If friction is present, how would the conservation-of-energy equation be modified? (e) Find the maximum speed of the child when the slide is frictionless if the initial height of the slide is 12.0 m.
In each situation shown in Figure P8.12, a ball moves from point A to point B. Use the following data to find the change in the gravitational potential energy in each case. You can assume that the radius of the ball is negligible. a. h = 1.35 m, = 25, and m = 0.65 kg b. R = 33.5 m and m = 756 kg c. R = 33.5 m and m = 756 kg FIGURE P8.12 Problems 12, 13, and 14.
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 Earths perihelion distance (closest approach to the Sun) is rp = 1.48 1011 m, and its aphelion distance (farthest point) is rA = 1.52 1011 m. What is the change in the SunEarths gravitational potential energy as the Earth moves from aphelion to perihelion? What is the change in its gravitational potential energy from perihelion to aphelion?
At the start of a basketball game, a referee tosses a basketball straight into the air by giving it some initial speed. After being given that speed, the ball reaches a maximum height ymax above where it started. Using conservation of energy, find expressions for a. the balls initial speed in terms of the gravitational acceleration g and the maximum height ymax and b. the height of the ball when it has speed v in terms of its current height y, the gravitational acceleration g, and the maximum height ymax.
A shopper weighs 3.00 kg of apples on a supermarket scale whose spring obeys Hookes law and notes that the spring stretches a distance of 3.00 cm. a. What will the springs extension be if 5.00 kg of oranges are weighed instead? b. What is the total amount of work that the shopper must do to stretch this spring a total distance of 7.00 cm beyond its relaxed position?
(a) Sketch a graph of the potential energy function U(x)=kx2/2+Aex2 where k , A, and are constants. (b) What is the force corresponding to this potential energy? (c) Suppose a particle of mass in moving with this potential energy has a velocity v when its position is x = . Show that the particle does not pass 2+2 through the origin unless Amv2=k22(1e a 2 ) .
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
The chin-up is one exercise that can be used to strengthen the biceps muscle. This muscle can exert a force of approximately 8.00 102 N as it contracts a distance of 7.5 cm in a 75-kg male.3 (a) How much work can the biceps muscles (one in each arm) perform in a single contraction? (b) Compare this amount of work with the energy required to lift a 75-kg person 40. cm in performing a chin-up. (c) Do you think the biceps muscle is the only muscle involved in performing a chin-up?
A block is hung from a vertical spring. The spring stretches (h = 0.0650 m) as shown for a particular instant in time in Figure P8.26. Consider the Earth, spring, and block to be in the system. If m = 0.865 kg and k = 125 N/m, find the change in the systems potential energy between the two times depicted in the figure. FIGURE P8.26
Suppose a horizontal force of 20 N is required to maintain a speed of 8 m/s of a 50 kg crate. (a) What is the power of this force? (b) Note that the acceleration of the crate is zero despite the fact that 20 N force acts on the crate horizontally. What happens to the energy given to the crate as a result of the work done by this 20 N force?