Question

You are using a rope to lift a 14.5 kg crate of fruit. Initially you are lifting the crate at 0.500 m/s. You then increase the tension in the rope to 190 N and lift the Grate an additional 1.15 m. During this d motion, how much work is done on the crate by the tension force? Express your answer with the appropriate units ! Å * OO? W Value Units Submit Request Answer Part D Find the work done by the force of gravity Express your answer with the appropriate units. HAO? W.. 0 Submit Previous Answers Request Answer x incorrect: Try Acain: 23 attempts remaining

Answer 1

Work done by any force is given by F. S = F*S*cos (theta) , theta is angle between force and displacement.
Work done by tension on crate = F.d = Fd coso F=190N, d = 1.15m, o=0. = Work done = 190x1.15X1 = 218. 5j Hence, Wook done on crate by tension Force is 218.5j Part-D. Work done by gravity = F.d = Ed coso Famg = 14.5*9.8 = 142) N ,dhism, 0 =180° work done = 142.1X1.18 X (-1) = -163.41 ung Hence, work done by Force of growity is -163.411 Part-E. Work done by Net Force = Freted - Freta coso. Fnet = T-mg = (190-142.1)N = 47.9N ---- () d = 1.15m Net Force is in the direction of displacement Hence, 0 = 0°. Coso = 1 bork done - 47.9 XUS XI 55.09 j. Part-F. initial velocity = 0.5oom/s Net Force = Tong = 47.9 [from i)] a = Net Force = 47.9 = 3.30 m/s² mass 14.5

Apply Formulae - v² = u²+2x axd v2 - (0.500)? + 2x 3-30 XMS = 0.25 +7.6 - 7.85 V = 57.85 m/s = 2.80 m/s kinetic Energy = £mv² = 1/2 x 14.587.85 = 56.91 ) Hence, the kinetic energy of the crate is 56.91 j.