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Question Number 216534 by Tawa11 last updated on 10/Feb/25

Commented by Tawa11 last updated on 10/Feb/25

In the figure, the block of mass M = 10kg rests on a spring located at the foot of an inclined plane that makes an angle θ = 30° with the horizontal. The spring force constant is k = 2,000 N/m. The spring is compressed at a distance of 10 cm (keeping the block resting on it) and is suddenly released. Assuming that the coefficient of friction between the plane and the block is μₖ = 0.2, calculate the total distance along the plane that the block travels before stopping momentarily.

Commented by Tawa11 last updated on 10/Feb/25

$$\mathrm{Is}\mathrm{this}\mathrm{formula}\mathrm{correct}\mathrm{sir}?$$$$\frac{1}{2}{\mathrm{ke}}_1^2-\mu \mathrm{R}\times 0.1=\mathrm{mgh}+\frac{1}{2}{\mathrm{ke}}_2^2$$

Commented by Tawa11 last updated on 10/Feb/25

Commented by Tawa11 last updated on 10/Feb/25

$$\mathrm{Is}\mathrm{this}\mathrm{solution}\mathrm{correct}?$$

Commented by mr W last updated on 10/Feb/25

$$wrong!$$$$\Rightarrow Q216110$$

Commented by Tawa11 last updated on 10/Feb/25

$$\mathrm{Sir},\mathrm{you}\mathrm{mean}\mathrm{the}\mathrm{formula}\mathrm{I}\mathrm{wrote}$$$$\mathrm{and}\mathrm{the}\mathrm{image}\mathrm{solution}\mathrm{is}\mathrm{wrong}.$$

Commented by mr W last updated on 10/Feb/25

$$both.$$

Commented by mr W last updated on 10/Feb/25

$$youcanstudyQ216110,thenyou$$$$knowwhythisiswrong.$$

Commented by Tawa11 last updated on 10/Feb/25

$$\mathrm{Am}\mathrm{still}\mathrm{trying}\mathrm{your}\mathrm{question}\mathrm{sir}.$$$$\mathrm{The}\mathrm{concept}\mathrm{is}\mathrm{giving}\mathrm{me}\mathrm{headache}$$

Commented by Tawa11 last updated on 10/Feb/25

$$\mathrm{Mechanics}\mathrm{is}\mathrm{hard}\mathrm{for}\mathrm{me}.$$$$\mathrm{But}\mathrm{I}\mathrm{will}\mathrm{not}\mathrm{give}\mathrm{up}$$

Commented by Tawa11 last updated on 10/Feb/25

$$\mathrm{I}\mathrm{thought}\mathrm{one}\mathrm{of}\mathrm{the}2\mathrm{will}\mathrm{be}\mathrm{right}.$$$$\mathrm{I}\mathrm{still}\mathrm{missed}\mathrm{it}.\mathrm{Sad}.$$

Commented by mr W last updated on 10/Feb/25

$$iftheblockisconnectedwiththe$$$$spring,thenboththeformulaand$$$$thesolutionarewrong.iftheblock$$$$isnotconnectedwiththespring,$$$$thentheformulaiswrong,butthe$$$$solutionisright.$$

Commented by Tawa11 last updated on 10/Feb/25

$$\mathrm{Sir},\mathrm{Q}216560,\mathrm{I}\mathrm{got}21.54\mathrm{m}/\mathrm{s}\mathrm{and}\mathrm{direction}201.8°$$

Commented by mr W last updated on 10/Feb/25

$${what}^{\prime }syouranswertothequestion$$$$above,ifM=20kg?$$

Commented by Tawa11 last updated on 10/Feb/25

$$\mathrm{Let}\mathrm{me}\mathrm{work}\mathrm{it}\mathrm{out}\mathrm{sir}.$$

Commented by Tawa11 last updated on 10/Feb/25

$$\mathrm{d}=\frac{10}{132.042}=0.076\mathrm{m}$$

Commented by mr W last updated on 10/Feb/25

$$thesameformulaapplied,butthis$$$$time{it}^{\prime }swrong!$$

Commented by Tawa11 last updated on 10/Feb/25

$$\mathrm{Haaaaaa}\mathrm{why}???$$$$\mathrm{physics}\mathrm{why}!!!!!!!.\mathrm{Why}\mathrm{sir}???$$

Commented by mr W last updated on 10/Feb/25

$$youassumedthatthespringwill$$$$berelaxed,i.e.d⩾0.1m.butwith$$$$d=0.075mitisnotrelaxed.that$$$$meansinthiscaseyoucannot$$$$assumethatthespringwillbe$$$$relaxed.$$

Commented by Tawa11 last updated on 10/Feb/25

$$\mathrm{Ohh}.$$$$\mathrm{Physics}\mathrm{thinking}\mathrm{is}\mathrm{wide}.$$

Commented by Tawa11 last updated on 10/Feb/25

$$\mathrm{Show}\mathrm{me}\mathrm{the}\mathrm{workings}\mathrm{sir}.$$$$\mathrm{Thanks}.$$

Commented by mr W last updated on 10/Feb/25

Commented by mr W last updated on 10/Feb/25

$$M=20kg$$$$\mathrm{\Delta }{l}_1=0.10m$$$$\mathrm{\Delta }{l}_2=\mathrm{\Delta }{l}_1-d$$$$\frac{k\mathrm{\Delta }{l}_1^2}{2}-d\mu Mg\cos \theta =\frac{k\mathrm{\Delta }{l}_2^2}{2}+Mgd\sin \theta$$$$\frac{k(\mathrm{\Delta }{l}_1^2-\mathrm{\Delta }{l}_2^2)}{2}=Mgd(\sin \theta +\mu \cos \theta )$$$$\frac{k(\mathrm{\Delta }{l}_1+\mathrm{\Delta }{l}_2)(\mathrm{\Delta }{l}_1-\mathrm{\Delta }{l}_2)}{2}=Mgd(\sin \theta +\mu \cos \theta )$$$$\frac{k(2\mathrm{\Delta }{l}_1-d)d}{2}=Mgd(\sin \theta +\mu \cos \theta )$$$$\frac{k(2\mathrm{\Delta }{l}_1-d)}{2}=Mg(\sin \theta +\mu \cos \theta )$$$$\Rightarrow d=2[\mathrm{\Delta }{l}_1-\frac{Mg(\sin \theta +\mu \cos \theta )}{k}]$$$$=2[0.1-\frac{20\times 9.81\times (1+0.2\times \sqrt{3})}{2\times 2000}]$$$$\approx 0.068m<0.10m✓$$

Commented by Tawa11 last updated on 10/Feb/25

$$\mathrm{Wow},\mathrm{thanks}\mathrm{for}\mathrm{your}\mathrm{time}\mathrm{sir}.$$

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