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 Lauren_6408 #1 Posted : Thursday, June 17, 2021 8:04:47 AM Rank: MemberGroups: Registered Joined: 5/16/2021Posts: 15Thanks: 0 timesWas thanked: 0 time(s) in 0 post(s) Hi there,I was wondering if someone could explain why there is no need to include the angle in this calculation of work. I thought forces perpendicular to the displacement do not do any work. In this question then I thought W = Fdsin(60). Because the object is moving vertically rather than horizontal I chose sin instead of cos. (Not sure if the picture will show up).Thanks Back to top User Profile
 INSTR_Katrina_128 #2 Posted : Tuesday, June 22, 2021 8:35:42 PM Rank: Advanced MemberGroups: Registered Joined: 5/18/2021Posts: 73Thanks: 0 timesWas thanked: 0 time(s) in 0 post(s) OOOH this question is evil!!! ðŸ˜«In this case, when you are lifting the 4 kg mass, you are doing work AGAINST GRAVITY, so the distance being travelled by the mass for the purposes of the work calculation is 5 m directly up.So, W = Fd = (4 kg)(10 m/s²)(5 m) = 200 J.Therefore, D is the best answer to the question.The angle has nothing to do with it because gravity is just straight up and down. ðŸ˜£ Back to top User Profile
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