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Question Number 216110 by Tawa11 last updated on 27/Jan/25

Commented by mr W last updated on 28/Jan/25

$$thisworkingsisok,ifwe{don}^{\prime }tthink$$$$abouthowtheparticlemovesfrom$$$$PtoQ.$$

Commented by mr W last updated on 28/Jan/25

Commented by Tawa11 last updated on 27/Jan/25

$$\mathrm{Hahahahaha}.$$$$\mathrm{Please}\mathrm{solve}\mathrm{with}\mathrm{the}240\mathrm{sir}.$$

Commented by Tawa11 last updated on 27/Jan/25

Commented by Tawa11 last updated on 27/Jan/25

Commented by Tawa11 last updated on 27/Jan/25

$$\mathrm{See}\mathrm{the}\mathrm{workings}\mathrm{I}\mathrm{saw}\mathrm{sir}.$$$$\mathrm{I}\mathrm{want}\mathrm{your}\mathrm{workings}\mathrm{sir}.$$

Commented by Tawa11 last updated on 27/Jan/25

$$\mathrm{Probably},\mathrm{he}\mathrm{meant}240\mathrm{GPa}.$$$$\mathrm{Even}\mathrm{this}\mathrm{is}\mathrm{out}\mathrm{of}\mathrm{range}.$$$$\mathrm{You}\mathrm{can}\mathrm{just}\mathrm{take}\mathrm{it}\mathrm{to}\mathrm{be}240\mathrm{sir}.$$$$\mathrm{I}\mathrm{need}\mathrm{your}\mathrm{approach}\mathrm{sir}.$$

Commented by Tawa11 last updated on 28/Jan/25

$$\mathrm{Sir},\mathrm{please}\mathrm{use}\mathrm{your}\mathrm{approach}\mathrm{for}\mathrm{both}\mathrm{cases}.$$$$\mathrm{without}\mathrm{considering}\mathrm{how}\mathrm{it}\mathrm{moves}\mathrm{and}\mathrm{also}$$$$\mathrm{consider}\mathrm{how}\mathrm{it}\mathrm{moves}.$$$$\mathrm{It}\mathrm{will}\mathrm{help}\mathrm{me}\mathrm{to}\mathrm{solve}\mathrm{some}\mathrm{other}\mathrm{exercises}\mathrm{by}\mathrm{myself}.$$$$\mathrm{thanks}\mathrm{sir}.$$

Commented by Tawa11 last updated on 28/Jan/25

$$\mathrm{I}\mathrm{usually}\mathrm{understand}\mathrm{your}\mathrm{approach}$$$$\mathrm{more}\mathrm{sir}.\mathrm{Thanks}\mathrm{for}\mathrm{being}\mathrm{there}\mathrm{always}.$$

Commented by mr W last updated on 28/Jan/25

$$wemustignorehowtheparticle$$$$movesfromPtoQ,otherwisewe$$$$cannotsolve.thatmeansthisis$$$$onlyaquestionintheoreticalphysics$$$$world,notinrealworld.$$$$asihavesaidtheworkingsaboveis$$$$ok.butitonlygivestheacceleration$$$$oftheparticleasitisleaving$$$$thepointC.buttheaccelerationof$$$$theparticleasitisapproachingthe$$$$pointCisdifferent.why?asthe$$$$particleisapprochingthepointC,$$$$thedirectionofthefrictionforceis$$$$downwards,whileasitisleaving$$$$thepoint,thedirectionoffriction$$$$isupwards.$$

Commented by Tawa11 last updated on 28/Jan/25

$$\mathrm{Sir},\mathrm{the},$$$$\mathrm{k}=\frac{240\mathrm{N}}{1.5\mathrm{m}}=160\mathrm{N}/\mathrm{m}?$$$$\mathrm{you}\mathrm{wrote}160\mathrm{m}/\mathrm{s}\mathrm{sir}.$$

Commented by mr W last updated on 28/Jan/25

$$k=160N/m=springconstant$$

Answered by mr W last updated on 28/Jan/25

$$k=\frac{240}{1.5}=160N/m$$$$\mathrm{\Delta }{l}_1=4-1.5=2.5m$$$$say\mid BC\mid =x$$$$\mathrm{\Delta }{l}_2=x-1.5$$$$energyconservation:$$$$\frac{k\mathrm{\Delta }{l}_1^2}{2}-\mu mg\cos \theta (x+4)=\frac{k\mathrm{\Delta }{l}_2^2}{2}+mg(x+4)\sin \theta$$$$\frac{k(\mathrm{\Delta }{l}_1^2-\mathrm{\Delta }{l}_2^2)}{2}=mg(x+4)(\sin \theta +\mu \cos \theta )$$$$\frac{160(2.5^2-{(x-1.5)}^2)}{2}=5\times 9.81(x+4)(0.8+0.5\times 0.6)$$$$16x^2-37.209x-20.836=0$$$$\Rightarrow x=\frac{37.209+\sqrt{37.{209}^2+4\times 16\times 20.836}}{2\times 16}$$$$\approx 2.792m$$$$\mathrm{\Delta }{l}_2=2.792-1.5=1.292m$$$$$$$$accelerationatpointC\left(before\right):$$$$a_{c1}=\frac{k\mathrm{\Delta }{l}_2+mg\sin \theta +\mu mg\cos \theta }{m}(\leftarrow )$$$$=\frac{k\mathrm{\Delta }{l}_2}{m}+g(\sin \theta +\mu \cos \theta )$$$$=\frac{160\times 1.292}{5}+9.81(0.8+0.5\times 0.6)$$$$=52.13m/s^2✓$$$$accelerationatpointC\left(after\right):$$$$a_{c2}=\frac{k\mathrm{\Delta }{l}_2+mg\sin \theta -\mu mg\cos \theta }{m}(\leftarrow )$$$$=\frac{k\mathrm{\Delta }{l}_2}{m}+g(\sin \theta -\mu \cos \theta )$$$$=\frac{160\times 1.292}{5}+9.81(0.8-0.5\times 0.6)$$$$=46.25m/s^2✓$$

Commented by mr W last updated on 28/Jan/25

Commented by mr W last updated on 28/Jan/25

Commented by Tawa11 last updated on 28/Jan/25

$$\mathrm{Wow},\mathrm{I}\mathrm{will}\mathrm{study}\mathrm{your}\mathrm{approach}\mathrm{sir}.$$$$\mathrm{And}\mathrm{moreover},\mathrm{his}\mathrm{answer}\mathrm{is}\mathrm{not}\mathrm{the}$$$$\mathrm{same}\mathrm{as}\mathrm{yours}\mathrm{sir}.$$$$\mathrm{He}\mathrm{got}46.3{\mathrm{ms}}^{-2}$$

Commented by Tawa11 last updated on 28/Jan/25

$$\mathrm{Ohh},\mathrm{probably}\mathrm{because}\mathrm{you}\mathrm{took}\mathrm{g}=10\mathrm{m}/{\mathrm{s}}^2.$$$$\mathrm{I}\mathrm{will}\mathrm{use}9.81\mathrm{m}/{\mathrm{s}}^2\mathrm{while}\mathrm{studying}\mathrm{your}\mathrm{solution}.$$

Commented by mr W last updated on 28/Jan/25

$$heandI{don}^{\prime }thavedifferent$$$$approach,butexactlythesame$$$$approach!$$

Commented by Tawa11 last updated on 28/Jan/25

$$\mathrm{But}\mathrm{I}\mathrm{understand}\mathrm{your}\mathrm{presentation}$$$$\mathrm{more}\mathrm{sir}.$$

Commented by mr W last updated on 28/Jan/25

$$nowyoushouldbeabletosolve$$$$followingadditionalquestion:$$$$atpointCtheparticlecomesto$$$$instantaneousrestforthefirst$$$$time.sayatpointDitcomesto$$$$instantaneousrestagain.find$$$$theaccelerationoftheparticleat$$$$pointD.$$

Commented by Tawa11 last updated on 28/Jan/25

$$\mathrm{I}\mathrm{will}\mathrm{work}\mathrm{on}\mathrm{it}\mathrm{sir}.$$

Commented by mr W last updated on 29/Jan/25

$${what}^{\prime }syourresult?$$

Commented by Tawa11 last updated on 29/Jan/25

$$\mathrm{Please}\mathrm{give}\mathrm{me}\mathrm{some}\mathrm{time}\mathrm{sir},$$$$\mathrm{I}\mathrm{am}\mathrm{having}\mathrm{exam},\mathrm{so},\mathrm{I}\mathrm{have}\mathrm{many}$$$$\mathrm{things}\mathrm{to}\mathrm{study}.\mathrm{Thanks}\mathrm{for}\mathrm{helping}$$$$\mathrm{me}\mathrm{sir}.$$

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