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 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

$$wonderful!$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com