Chapter 8. Work, Energy, and Power
Work
8-1. What is the work done by a force of 20 N acting through a parallel distance of 8 m? What force will do the same work through a distance of 4 m?
Work = (20 N)(8 m) = 160 J ; F (4 m) = 160 J; F = 40.0 N
8-2. A worker lifts a 40 lb weight through a height of 10 ft. How many meters can a 10-kg block be lifted by the same amount ofwork?
Work = (20 lb)(10 ft) = 200 ft lb; [pic]
Work = Fs = mgs; [pic]; s = 2.77 m
8-3. A tugboat exerts a constant force of 4000 N on a ship, moving it a distance of 15 m. What work is done?
Work = (4000 N)(15 m); Work = 60,000 J
8-4. A 5-kg hammer is lifted to a height of 3 m. What is the minimum required work?
Work = Fs = (5 kg)(9.8 m/s2)(3 m); Work = 147 J8-5. A push of 30 lb is applied along the handle of a lawn mower producing a horizontal displacement of 40 ft. If the handle makes an angle of 300 with the ground, what work was done by the 30-lb force?
Work = (F cos ()s = (30 lb) cos 300 (40 ft)
Work = 1040 ft lb
8-6. The trunk in Fig. 8-10 is dragged a horizontal distanceof 24 m by a rope that makes an angle ( with the floor. If the rope tension is 8 N, what works are done for the following angles: 00, 300, 600, 900?
Work = (F cos ()s = (8 N) cos 00 (24 m) = 192 J
Work = (8 N) cos 300 (24 m) = 166 J ; Work60 = 96 J ; Work90 = 0 J
8-7. A horizontal force pushes a 10-kg sled along a driveway for a distance of 40 m. If thecoefficient of sliding friction is 0.2, what work is done by the friction force?
Work = (F cos ()s = (F) (cos 1800)s = – F s; but F = (kN = (k mg
Work = (kmg s = (0.2)(10 kg)(9.8 m/s2)(40 m); Work = –784 J
*8-8. A sled is dragged a distance of 12.0 m by a rope under constant tension of 140 N. The task requires 1200 J of work. What angle does the rope make with the ground?
Work =(F cos ()s; [pic]
cos ( = 0.714; ( = 44.40
Resultant Work
8-9. An average force of 40 N compresses a coiled spring a distance of 6 cm. What is the work done by the 40-N force? What work done by the spring? What is the resultant work? Work40 = (40 N)(0.06 m) = 2.40 J, (positive work)
Worksp = (-40 N)(0.06 m) = -2.40 J, (negativework)
Resultant work = ?(works) = 2.4 J – 2.4 J = 0 J
Work is positive when force is with displacement, negative when against displacement.
8-10. A horizontal force of 20 N drags a small sled 42 m across the ice at constant speed. Find the work done by the pulling force and by the friction force. What is the resultant force?
Work40 = (20 N)(24 m) = 2.40 J,(positive work)
Worksp = (-20 N)(24 m) = -2.40 J, (negative work)
Resultant force and, hence, resultant work are zero.
*8-11. A 10-kg block is dragged 20 m by a parallel force of 26 N. If (k = 0.2, what is the resultant work and what acceleration results.
F = (kN = (kmg F = 0.2(10 kg)(9.8 m/s2) = 19.6 N
Work = FR s = (P – F)s; Work = (26 N – 19.6 N)(20 m)Work = 128 J
FR = (26 N – 19.6 N) = 6.40 N; [pic]; a = 0.640 m/s2
*8-12. A rope making an angle of 350 drags a 10-kg toolbox a horizontal distance of 20 m. The tension in the rope is 60 N and the constant friction force is 30 N. What work is done by the rope? What work is done by friction? What is the resultant work?
(Work)rope = (60 N) cos 350 (20 m);(Work)r = 983 J
(Work)F = (-30 N)(20 m) = -600 J; (Work)F = -600 J
Resultant Work = ?(works) = 983 J – 600 J; Resultant Work = 383 J
Extra work can show that for this example, ( k = 0.472
*8-13. For the example described in Problem 8-12, what is the coefficient of friction between the toolbox and the floor. (Refer to figure and information given in…