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Question-216110




Question Number 216110 by Tawa11 last updated on 27/Jan/25
Commented by mr W last updated on 28/Jan/25
this workings is ok, if we don′t think  about how the particle moves from  P  to Q.
thisworkingsisok,ifwedontthinkabouthowtheparticlemovesfromPtoQ.
Commented by mr W last updated on 28/Jan/25
Commented by Tawa11 last updated on 27/Jan/25
Hahahahaha.  Please solve with the 240 sir.
Hahahahaha.Pleasesolvewiththe240sir.
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
See the workings I saw sir.  I want your workings sir.
SeetheworkingsIsawsir.Iwantyourworkingssir.
Commented by Tawa11 last updated on 27/Jan/25
Probably, he meant  240GPa.  Even this is out of range.  You can just take it to be  240 sir.  I need your approach sir.
Probably,hemeant240GPa.Eventhisisoutofrange.Youcanjusttakeittobe240sir.Ineedyourapproachsir.
Commented by Tawa11 last updated on 28/Jan/25
Sir, please use your approach for both cases.  without considering how it moves and also  consider how it moves.  It will help me to solve some other exercises by myself.  thanks sir.
Sir,pleaseuseyourapproachforbothcases.withoutconsideringhowitmovesandalsoconsiderhowitmoves.Itwillhelpmetosolvesomeotherexercisesbymyself.thankssir.
Commented by Tawa11 last updated on 28/Jan/25
I usually understand your approach  more sir. Thanks for being there always.
Iusuallyunderstandyourapproachmoresir.Thanksforbeingtherealways.
Commented by mr W last updated on 28/Jan/25
we must ignore how the particle  moves from P to Q, otherwise we  can not solve. that means this is  only a question in theoretical physics  world, not in real world.  as i have said the workings above is  ok. but it only gives the acceleration  of the particle as it is leaving  the point C. but the acceleration of   the particle as it is approaching the  point C is different. why? as the  particle is approching the point C,  the direction of thefriction force is  downwards, while as it is leaving  the point, the direction of friction  is upwards.
wemustignorehowtheparticlemovesfromPtoQ,otherwisewecannotsolve.thatmeansthisisonlyaquestionintheoreticalphysicsworld,notinrealworld.asihavesaidtheworkingsaboveisok.butitonlygivestheaccelerationoftheparticleasitisleavingthepointC.buttheaccelerationoftheparticleasitisapproachingthepointCisdifferent.why?astheparticleisapprochingthepointC,thedirectionofthefrictionforceisdownwards,whileasitisleavingthepoint,thedirectionoffrictionisupwards.
Commented by Tawa11 last updated on 28/Jan/25
Sir, the,    k  =  ((240N)/(1.5m))   =  160N/m?  you wrote   160 m/s  sir.
Sir,the,k=240N1.5m=160N/m?youwrote160m/ssir.
Commented by mr W last updated on 28/Jan/25
k=160 N/m =spring constant
k=160N/m=springconstant
Answered by mr W last updated on 28/Jan/25
k=((240)/(1.5))=160 N/m  Δl_1 =4−1.5=2.5 m  say ∣BC∣=x  Δl_2 =x−1.5  energy conservation:  ((kΔl_1 ^2 )/2)−μmg cos θ (x+4)=((kΔl_2 ^2 )/2)+mg(x+4) sin θ  ((k(Δl_1 ^2 −Δl_2 ^2 ))/2)=mg(x+4)(sin θ+μ cos θ)  ((160(2.5^2 −(x−1.5)^2 ))/2)=5×9.81(x+4)(0.8+0.5×0.6)  16x^2 −37.209x−20.836=0  ⇒x=((37.209+(√(37.209^2 +4×16×20.836)))/(2×16))         ≈2.792 m  Δl_2 =2.792−1.5=1.292 m    acceleration at point C (before):  a_(c1) =((kΔl_2 +mg sin θ+μmg cos θ)/m)  (←)        =((kΔl_2 )/m)+g(sin θ+μ cos θ)        =((160×1.292)/5)+9.81(0.8+0.5×0.6)        =52.13 m/s^2    ✓  acceleration at point C (after):  a_(c2) =((kΔl_2 +mg sin θ−μmg cos θ)/m)  (←)        =((kΔl_2 )/m)+g(sin θ−μ cos θ)        =((160×1.292)/5)+9.81(0.8−0.5×0.6)        =46.25 m/s^2    ✓
k=2401.5=160N/mΔl1=41.5=2.5msayBC∣=xΔl2=x1.5energyconservation:kΔl122μmgcosθ(x+4)=kΔl222+mg(x+4)sinθk(Δl12Δl22)2=mg(x+4)(sinθ+μcosθ)160(2.52(x1.5)2)2=5×9.81(x+4)(0.8+0.5×0.6)16x237.209x20.836=0x=37.209+37.2092+4×16×20.8362×162.792mΔl2=2.7921.5=1.292maccelerationatpointC(before):ac1=kΔl2+mgsinθ+μmgcosθm()=kΔl2m+g(sinθ+μcosθ)=160×1.2925+9.81(0.8+0.5×0.6)=52.13m/s2accelerationatpointC(after):ac2=kΔl2+mgsinθμmgcosθm()=kΔl2m+g(sinθμcosθ)=160×1.2925+9.81(0.80.5×0.6)=46.25m/s2
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
Wow, I will study your approach sir.  And moreover, his answer is not the  same as yours sir.  He got  46.3 ms^(−2)
Wow,Iwillstudyyourapproachsir.Andmoreover,hisanswerisnotthesameasyourssir.Hegot46.3ms2
Commented by Tawa11 last updated on 28/Jan/25
Ohh, probably because you took  g  =  10m/s^2 .  I will use  9.81 m/s^2  while studying your solution.
Ohh,probablybecauseyoutookg=10m/s2.Iwilluse9.81m/s2whilestudyingyoursolution.
Commented by mr W last updated on 28/Jan/25
he and I don′t have different  approach, but exactly the same   approach!
heandIdonthavedifferentapproach,butexactlythesameapproach!
Commented by Tawa11 last updated on 28/Jan/25
But I understand your presentation  more sir.
ButIunderstandyourpresentationmoresir.
Commented by mr W last updated on 28/Jan/25
now you should be able to solve  following additional question:  at point C the particle comes to   instantaneous rest for the first  time. say at point D it comes to  instantaneous rest again. find  the acceleration of the particle at  point D.
nowyoushouldbeabletosolvefollowingadditionalquestion:atpointCtheparticlecomestoinstantaneousrestforthefirsttime.sayatpointDitcomestoinstantaneousrestagain.findtheaccelerationoftheparticleatpointD.
Commented by Tawa11 last updated on 28/Jan/25
I will work on it sir.
Iwillworkonitsir.
Commented by mr W last updated on 29/Jan/25
what′s your result?
whatsyourresult?
Commented by Tawa11 last updated on 29/Jan/25
Please give me some time sir,  I am having exam, so, I have many  things to study. Thanks for helping  me sir.
Pleasegivemesometimesir,Iamhavingexam,so,Ihavemanythingstostudy.Thanksforhelpingmesir.

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