Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

Question Number 20986 by Tinkutara last updated on 09/Sep/17

A 50 kg log rest on the smooth horizontal  surface. A motor deliver a towing force  T as shown below. The momentum of  the particle at t = 5 s is

$$\mathrm{A}\:\mathrm{50}\:\mathrm{kg}\:\mathrm{log}\:\mathrm{rest}\:\mathrm{on}\:\mathrm{the}\:\mathrm{smooth}\:\mathrm{horizontal} \\ $$$$\mathrm{surface}.\:\mathrm{A}\:\mathrm{motor}\:\mathrm{deliver}\:\mathrm{a}\:\mathrm{towing}\:\mathrm{force} \\ $$$${T}\:\mathrm{as}\:\mathrm{shown}\:\mathrm{below}.\:\mathrm{The}\:\mathrm{momentum}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{particle}\:\mathrm{at}\:{t}\:=\:\mathrm{5}\:\mathrm{s}\:\mathrm{is} \\ $$

Commented by Tinkutara last updated on 09/Sep/17

Answered by dioph last updated on 09/Sep/17

P(t) = ∫T(t)dt  P(5) = ∫_0 ^4 40t^2 dt + ∫_4 ^5 640dt  P(5) = ((40)/3)t^3 ∣_0 ^4  + 640t∣_4 ^5   P(5) = ((40×64)/3) + 640×1  P(5) ≅ 1493.33 kg m/s

$${P}\left({t}\right)\:=\:\int{T}\left({t}\right){dt} \\ $$$${P}\left(\mathrm{5}\right)\:=\:\int_{\mathrm{0}} ^{\mathrm{4}} \mathrm{40}{t}^{\mathrm{2}} {dt}\:+\:\int_{\mathrm{4}} ^{\mathrm{5}} \mathrm{640}{dt} \\ $$$${P}\left(\mathrm{5}\right)\:=\:\frac{\mathrm{40}}{\mathrm{3}}{t}^{\mathrm{3}} \mid_{\mathrm{0}} ^{\mathrm{4}} \:+\:\mathrm{640}{t}\mid_{\mathrm{4}} ^{\mathrm{5}} \\ $$$${P}\left(\mathrm{5}\right)\:=\:\frac{\mathrm{40}×\mathrm{64}}{\mathrm{3}}\:+\:\mathrm{640}×\mathrm{1} \\ $$$${P}\left(\mathrm{5}\right)\:\cong\:\mathrm{1493}.\mathrm{33}\:\mathrm{kg}\:\mathrm{m}/\mathrm{s} \\ $$

Commented by Tinkutara last updated on 10/Sep/17

Thank you very much Sir!

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com