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

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

Commented by Tinkutara last updated on 09/Sep/17

Answered by dioph last updated on 09/Sep/17

$$P\left(t\right)=\int T\left(t\right)dt$$$$P\left(5\right)={\int }_0^{4}40t^{2}dt+{\int }_4^{5}640dt$$$$P\left(5\right)=\frac{40}{3}{t}^3{\mid }_0^4+640t{\mid }_4^5$$$$P\left(5\right)=\frac{40\times 64}{3}+640\times 1$$$$P\left(5\right)\cong 1493.33\mathrm{kg}\mathrm{m}/\mathrm{s}$$

Commented by Tinkutara last updated on 10/Sep/17

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

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