a-A-body-of-mass-m-initially-at-rest-at-a-point-O-on-a-smooth-horizontal-surface-A-horizontal-force-F-is-applied-to-the-body-and-caused-it-to-move-in-a-straight-line-accross-the-surface-The-magnit

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Question Number 13302 by tawa tawa last updated on 17/May/17

$$\left(\mathrm{a}\right)$$$$\mathrm{A}\mathrm{body}\mathrm{of}\mathrm{mass}\mathrm{m}\mathrm{initially}\mathrm{at}\mathrm{rest}\mathrm{at}\mathrm{a}\mathrm{point}\mathrm{O}\mathrm{on}\mathrm{a}\mathrm{smooth}\mathrm{horizontal}\mathrm{surface}.$$$$\mathrm{A}\mathrm{horizontal}\mathrm{force}\mathrm{F}\mathrm{is}\mathrm{applied}\mathrm{to}\mathrm{the}\mathrm{body}\mathrm{and}\mathrm{caused}\mathrm{it}\mathrm{to}\mathrm{move}\mathrm{in}\mathrm{a}\mathrm{straight}$$$$\mathrm{line}\mathrm{accross}\mathrm{the}\mathrm{surface}.\mathrm{The}\mathrm{magnitude}\mathrm{of}\mathrm{F}\mathrm{is}\mathrm{given}\mathrm{by}\mathrm{F}=\frac{1}{\mathrm{s}+\alpha },\mathrm{Where}\mathrm{S}\mathrm{is}$$$$\mathrm{the}\mathrm{distance}\mathrm{of}\mathrm{the}\mathrm{body}\mathrm{from}\mathrm{O}\mathrm{and}\alpha \mathrm{is}\mathrm{the}\mathrm{positive}\mathrm{constant}.\mathrm{If}\mathrm{v}\mathrm{is}\mathrm{the}\mathrm{speed}$$$$\mathrm{of}\mathrm{the}\mathrm{body}\mathrm{at}\mathrm{any}\mathrm{moment},\mathrm{Show}\mathrm{that}\mathrm{S}=\alpha ({\mathrm{e}}^{\frac{1}{2}{\mathrm{mv}}^2}-1).$$$$\left(\mathrm{b}\right)$$$$\mathrm{If}\mathrm{F}=15\mathrm{s}+4\mathrm{and}\mathrm{m}=1\mathrm{kg}\mathrm{and}\mathrm{the}\mathrm{body}\mathrm{is}\mathrm{initially}\mathrm{at}\mathrm{rest}\mathrm{at}\mathrm{point}\mathrm{O}.$$$$\mathrm{Determine},$$$$\left(\mathrm{i}\right)\mathrm{v}\mathrm{when}\mathrm{s}=2\mathrm{m}\left(\mathrm{ii}\right)\mathrm{s}\mathrm{when}\mathrm{v}=8\mathrm{m}/\mathrm{s}$$

Answered by mrW1 last updated on 17/May/17

$$\left(a\right)$$$$F=ma$$$$a=\frac{dv}{dt}=\frac{dv}{ds}\times \frac{ds}{dt}=v\frac{dv}{ds}$$$$F=\frac{1}{s+\alpha }$$$$\Rightarrow \frac{1}{s+\alpha }=mv\frac{dv}{ds}$$$$\Rightarrow \frac{ds}{s+\alpha }=mvdv$$$$\Rightarrow \ln (s+\alpha )=\frac{1}{2}{mv}^2+C$$$$ats=0,v=0$$$$\Rightarrow \ln \alpha =C$$$$\Rightarrow \ln (s+\alpha )=\frac{1}{2}{mv}^2+\ln \alpha$$$$\Rightarrow \ln (s+\alpha )-\ln \alpha =\frac{1}{2}{mv}^2$$$$\Rightarrow \ln (\frac{s}{\alpha }+1)=\frac{1}{2}{mv}^2$$$$\Rightarrow \frac{s}{\alpha }+1=e^{\frac{1}{2}{mv}^2}$$$$\Rightarrow s=\alpha (e^{\frac{1}{2}{mv}^2}-1)$$$$$$$$\left(b\right)$$$$F=ma=mv\frac{dv}{ds}$$$$15s+4=v\frac{dv}{ds}$$$$(15s+4)ds=vdv$$$$\frac{1}{30}{(15s+4)}^2=\frac{1}{2}{v}^2+C$$$$ats=0,v=0$$$$\Rightarrow \frac{16}{30}=C$$$$\frac{1}{30}{(15s+4)}^2=\frac{1}{2}{v}^2+\frac{16}{30}$$$${(15s+4)}^2=15v^2+16$$$$15v^2={(15s+4)}^2-4^2=15s(15s+8)$$$$v^2=s(15s+8)$$$$\Rightarrow v=\sqrt{s(15s+8)}$$$$$$$$\left(i\right)$$$$s=2m\Rightarrow v=\sqrt{2(30+8)}=2\sqrt{19}=8.7m/s$$$$\left(ii\right)$$$$8=\sqrt{s(15s+8)}$$$$15s^2+8s-64=0$$$$s=\frac{-8+\sqrt{64+4\times 15\times 64}}{2\times 15}=\frac{-4+4\sqrt{61}}{15}=1.82m$$

Commented by tawa tawa last updated on 18/May/17

$$\mathrm{Wow},\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.\mathrm{I}\mathrm{really}\mathrm{appreciate}.$$

Answered by ajfour last updated on 18/May/17

$$\left(\mathit{a}\right)$$$$△\mathit{K}=\mathit{W}={\int }_0^s\overline{\mathit{F}}.d\overline{\mathit{s}}={\int }_0^s\frac{ds}{s+\alpha }$$$$\frac{1}{2}{\mathit{m}\mathit{v}}^2=\ln (1+\frac{\mathit{s}}{\mathit{\alpha }})$$$$\mathit{s}=\mathit{\alpha }({\mathit{e}}^{\frac{1}{2}{\mathit{m}\mathit{v}}^2}-1).$$$$$$$$\left(\mathit{b}\right)$$$$\frac{1}{2}{\mathit{m}\mathit{v}}^2={\int }_0^s(15\mathit{s}+4)d\mathit{s}$$$$=\frac{15}{2}{s}^2+4s$$$${\mathit{v}}^2=15{\mathit{s}}^2+8\mathit{s}$$$$for\mathit{m}=1kgand\mathit{s}=2m$$$${\mathit{v}}^2=15\left(4\right)+8\left(2\right)$$$$\mathit{v}=2\sqrt{19}m/s\approx 8.72m/s$$$$when\mathit{v}=8m/s$$$$64=15{\mathit{s}}^2+8\mathit{s}$$$$15{\mathit{s}}^2+8\mathit{s}-64=0$$$$\mathit{s}=\frac{-8+\sqrt{64+60\times 64}}{30}$$$$\mathit{s}=\frac{-8+8\sqrt{61}}{30}\approx 1.82m$$$$$$

Commented by tawa tawa last updated on 18/May/17

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.\mathrm{i}\mathrm{really}\mathrm{appreciate}.$$

Commented by mrW1 last updated on 18/May/17

$$wonderful!$$

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