HW3 Solutions

# B counting the areas under the axis as negative

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(b) Counting the ―areas‖ under the axis as negative contributions, we find (for 0 x 7) the work to be 30 J at x = 7.0 m. (c) And at x = 9.0 m, the work is 12 J. (d) Equation 7-10 (along with Eq. 7-1) leads to speed v = 6.5 m/s at x = 4.0 m. Returning to the original graph (where a was plotted) we note that (since it started from rest) it has received acceleration(s) (up to this point) only in the + x direction and consequently must have a velocity vector pointing in the + x direction at x = 4.0 m. (e) Now, using the result of part (b) and Eq. 7-10 (along with Eq. 7-1) we find the speed is 5.5 m/s at x = 7.0 m. Although it has experienced some deceleration during the 0 x 7 interval, its velocity vector still points in the + x direction. (f) Finally, using the result of part (c) and Eq. 7-10 (along with Eq. 7-1) we find its speed v = 3.5 m/s at x = 9.0 m. It certainly has experienced a significant amount of deceleration during the 0 x 9 interval; nonetheless, its velocity vector still points in the + x direction. 10. P. 7-38. (a) Using the work-kinetic energy theorem 2.0 2 3 0 1 (2.5 ) 0 (2.5)(2.0) (2.0) 2.3 J. 3 f i K K x dx

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(b) For a variable end-point, we have K f as a function of x , which could be differentiated to find the extremum value, but we recognize that this is equivalent to solving F = 0 for x : 2 0 2.5 0 F x . Thus, K is extremized at 2.5 1.6 m x and we obtain 2.5 2 3 0 1 (2.5 ) 0 (2.5)( 2.5) ( 2.5) 2.6 J. 3 f i K K x dx Recalling our answer for part (a), it is clear that this extreme value is a maximum. 11. P.7-49. We have a loaded elevator moving upward at a constant speed. The forces involved are: gravitational force on the elevator, gravitational force on the counterweight, and the force by the motor via cable. The total work is the sum of the work done by gravity on the elevator, the work done by gravity on the counterweight, and the work done by the motor on the system: e c m W W W W . Since the elevator moves at constant velocity, its kinetic energy does not change and according to the work-kinetic energy theorem the total work done is zero, that is, 0 W K   . The elevator moves upward through 54 m, so the work done by gravity on it is 2 5 (1200 kg)(9.80 m/s )(54 m) 6.35 10 J. e e W m gd    The counterweight moves downward the same distance, so the work done by gravity on it is 2 5 (950 kg)(9.80 m/s )(54 m) 5.03 10 J. c c W m gd Since W = 0, the work done by the motor on the system is 5 5 5 6.35 10 J 5.03 10 J 1.32 10 J. m e c W W W  This work is done in a time interval of 3.0 min 180 s, t   so the power supplied by the motor to lift the elevator is 5 2 1.32 10 J 7.4 10 W. 180 s m W P t 12. P. 7-51. (a) The object’s displacement is ˆ ˆ ˆ ( 8.00 m)i (6.00 m)j (2.00 m)k . f i d d d   Thus, Eq. 7-8 gives
(3.00 N)( 8.00 m) (7.00 N)(6.00 m) (7.00 N)(2.00 m) 32.0 J. W F d (b) The average power is given by Eq. 7-42: avg 32.0 8.00 W.

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