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Introductory Physics Homework Help
Total work done while pushing a wheelchair
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[QUOTE="TheBigDig, post: 6564070, member: 587059"] [B]Homework Statement:[/B] A person in a wheelchair (a total mass of 100 kg) is pushed from rest to a speed of 2.0 m/s . If the effective coefficient of friction is 0.05 and the distance moved is 15 m, what is the total work done? [B]Relevant Equations:[/B] ##v^2 = u^2 +2as## ## F = ma ## ##f = \mu_k N## ##W = Fs## [ATTACH type="full" alt="2021_11_14 14_18 Office Lens.jpg"]292266[/ATTACH] Taking v = 2m/s, u=0m/s and s = 15m, we get ##a=0.13m/s^2## ##F_g = mg = 100(9.8) = 980N## Since there's no vertical acceleration, the normal force is equal to the weight ##N = 980N## ##f = \mu_k N = 0.05(980) = 49N## ##F_{net} = ma = 100(0.13) = 13N## ##F_{app} = F_{net}+f = 62N## My question is, is my work done the applied force by the distance or the net force by the distance. I'm a little unsure of it. [/QUOTE]
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Total work done while pushing a wheelchair
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