Bộ tài liệu bài tập tham khảo môn vật lý bậc đại học bằng tiếng anh Chapter 06 | a b N c J - . a Pulling slowly can be taken to mean that the bucket rises at constant speed so the tension in the rope may be taken to be the bucket s weight. In pulling a given length of rope from Eq. W Fs mgs kg m s2 m J. b Gravity is directed opposite to the direction of the bucket s motion so Eq. gives the negative of the result of part a or - 265 J . c The net work done on the bucket is zero. 300J. a The friction force to be overcome is f Mk n juk mg s2 N or 74 N to two figures. b From Eq. Fs N m 331J . The work is positive since the worker is pushing in the same direction as the crate s motion. c Since f and s are oppositely directed Eq. gives - fs - N m -331J. d Both the normal force and gravity act perpendicular to the direction of motion so neither force does work. e The net work done is zero. a See Exercise . The needed force is r mg 30kg s2 F --------------- ------------------------- N cos 0- ksin cos30 - sin30 keeping extra figures. b Fscos0 N cos 30 J again keeping an extra figure. c The normal force is mg F sin0 and so the work done by friction is - m 30 kg m s2 sin30 . d Both the normal force and gravity act perpendicular to the direction of motion so neither force does work. e The net work done is zero. From Eq. Fscos b 180N 300 m x 104 J. 2Fscos0 2 x 106 N x 103 m cos14 x 109 J or x 109 J to two places. The work you do is F- i 30N i - 40N j i - j 30N -40N -270N m 120N m -150J a i Tension force is always perpendicular to the displacement and does no work. ii Work done by gravity is - mg y2 - yj. When y y2 Wmg 0 . b i Tension does no work. ii Let lbe the length of the string. Wmg -mg y2 -y1 -mg 2 J The displacement is upward and the .
