Question-216534

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

Is this formula correct sir? (1/2)ke_1 ^2 − μR × 0.1 = mgh + (1/2)ke_2 ^2

$$\mathrm{Is}\:\mathrm{this}\:\mathrm{formula}\:\mathrm{correct}\:\mathrm{sir}? \\ $$$$\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ke}_{\mathrm{1}} ^{\mathrm{2}} \:\:−\:\:\mu\mathrm{R}\:×\:\mathrm{0}.\mathrm{1}\:\:\:=\:\:\:\mathrm{mgh}\:\:+\:\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ke}_{\mathrm{2}} ^{\mathrm{2}} \\ $$

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

$${wrong}! \\ $$$$\Rightarrow\:{Q}\mathrm{216110} \\ $$

Commented by Tawa11 last updated on 10/Feb/25

Sir, you mean the formula I wrote and the image solution is wrong.

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

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

$${both}. \\ $$

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

you can study Q216110, then you know why this is wrong.

$${you}\:{can}\:{study}\:{Q}\mathrm{216110},\:{then}\:{you} \\ $$$${know}\:{why}\:{this}\:{is}\:{wrong}. \\ $$

Commented by Tawa11 last updated on 10/Feb/25

Am still trying your question sir. The concept is giving me headache

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

Commented by Tawa11 last updated on 10/Feb/25

Mechanics is hard for me. But I will not give up

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

Commented by Tawa11 last updated on 10/Feb/25

I thought one of the 2 will be right. I still missed it. Sad.

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

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

$${if}\:{the}\:{block}\:{is}\:{connected}\:{with}\:{the} \\ $$$${spring},\:{then}\:{both}\:{the}\:{formula}\:{and} \\ $$$${the}\:{solution}\:{are}\:{wrong}.\:{if}\:{the}\:{block} \\ $$$${is}\:{not}\:{connected}\:{with}\:{the}\:{spring}, \\ $$$${then}\:{the}\:{formula}\:{is}\:{wrong},\:{but}\:{the} \\ $$$${solution}\:{is}\:{right}. \\ $$

Commented by Tawa11 last updated on 10/Feb/25

Sir, Q216560, I got 21.54m/s and direction 201.8°

$$\mathrm{Sir},\:\:\mathrm{Q216560},\:\:\mathrm{I}\:\mathrm{got}\:\:\mathrm{21}.\mathrm{54m}/\mathrm{s}\:\:\mathrm{and}\:\mathrm{direction}\:\mathrm{201}.\mathrm{8}° \\ $$

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

what′s your answer to the question above, if M=20kg?

$${what}'{s}\:{your}\:{answer}\:{to}\:{the}\:{question} \\ $$$${above},\:{if}\:{M}=\mathrm{20}{kg}? \\ $$

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

$$\mathrm{d}\:\:=\:\:\frac{\mathrm{10}}{\mathrm{132}.\mathrm{042}}\:\:=\:\:\mathrm{0}.\mathrm{076m} \\ $$

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

$${the}\:{same}\:{formula}\:{applied},\:{but}\:{this}\: \\ $$$${time}\:{it}'{s}\:{wrong}! \\ $$

Commented by Tawa11 last updated on 10/Feb/25

Haaaaaa why??? physics why!!!!!!!. Why sir???

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

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

you assumed that the spring will be relaxed, i.e. d≥0.1m. but with d=0.075m it is not relaxed. that means in this case you can not assume that the spring will be relaxed.

$${you}\:{assumed}\:{that}\:{the}\:{spring}\:{will} \\ $$$${be}\:{relaxed},\:{i}.{e}.\:{d}\geqslant\mathrm{0}.\mathrm{1}{m}.\:\:{but}\:{with}\: \\ $$$${d}=\mathrm{0}.\mathrm{075}{m}\:{it}\:{is}\:{not}\:{relaxed}.\:{that}\: \\ $$$${means}\:{in}\:{this}\:{case}\:{you}\:{can}\:{not}\: \\ $$$${assume}\:{that}\:{the}\:{spring}\:{will}\:{be}\: \\ $$$${relaxed}. \\ $$

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=20 kg Δl_1 =0.10m Δl_2 =Δl_1 −d ((kΔl_1 ^2 )/2)−dμMg cos θ=((kΔl_2 ^2 )/2)+Mgd sin θ ((k(Δl_1 ^2 −Δl_2 ^2 ))/2)=Mgd (sin θ+μ cos θ) ((k(Δl_1 +Δl_2 )(Δl_1 −Δl_2 ))/2)=Mgd (sin θ+μ cos θ) ((k(2Δl_1 −d)d)/2)=Mgd (sin θ+μ cos θ) ((k(2Δl_1 −d))/2)=Mg (sin θ+μ cos θ) ⇒d=2[Δl_1 −((Mg(sin θ+μ cos θ))/k)] =2[0.1−((20×9.81×(1+0.2×(√3)))/(2×2000))] ≈0.068 m <0.10 m ✓

$${M}=\mathrm{20}\:{kg} \\ $$$$\Delta{l}_{\mathrm{1}} =\mathrm{0}.\mathrm{10}{m} \\ $$$$\Delta{l}_{\mathrm{2}} =\Delta{l}_{\mathrm{1}} −{d} \\ $$$$\frac{{k}\Delta{l}_{\mathrm{1}} ^{\mathrm{2}} }{\mathrm{2}}−{d}\mu{Mg}\:\mathrm{cos}\:\theta=\frac{{k}\Delta{l}_{\mathrm{2}} ^{\mathrm{2}} }{\mathrm{2}}+{Mgd}\:\mathrm{sin}\:\theta \\ $$$$\frac{{k}\left(\Delta{l}_{\mathrm{1}} ^{\mathrm{2}} −\Delta{l}_{\mathrm{2}} ^{\mathrm{2}} \right)}{\mathrm{2}}={Mgd}\:\left(\mathrm{sin}\:\theta+\mu\:\mathrm{cos}\:\theta\right) \\ $$$$\frac{{k}\left(\Delta{l}_{\mathrm{1}} +\Delta{l}_{\mathrm{2}} \right)\left(\Delta{l}_{\mathrm{1}} −\Delta{l}_{\mathrm{2}} \right)}{\mathrm{2}}={Mgd}\:\left(\mathrm{sin}\:\theta+\mu\:\mathrm{cos}\:\theta\right) \\ $$$$\frac{{k}\left(\mathrm{2}\Delta{l}_{\mathrm{1}} −{d}\right){d}}{\mathrm{2}}={Mgd}\:\left(\mathrm{sin}\:\theta+\mu\:\mathrm{cos}\:\theta\right) \\ $$$$\frac{{k}\left(\mathrm{2}\Delta{l}_{\mathrm{1}} −{d}\right)}{\mathrm{2}}={Mg}\:\left(\mathrm{sin}\:\theta+\mu\:\mathrm{cos}\:\theta\right) \\ $$$$\Rightarrow{d}=\mathrm{2}\left[\Delta{l}_{\mathrm{1}} −\frac{{Mg}\left(\mathrm{sin}\:\theta+\mu\:\mathrm{cos}\:\theta\right)}{{k}}\right] \\ $$$$\:\:\:\:\:\:\:=\mathrm{2}\left[\mathrm{0}.\mathrm{1}−\frac{\mathrm{20}×\mathrm{9}.\mathrm{81}×\left(\mathrm{1}+\mathrm{0}.\mathrm{2}×\sqrt{\mathrm{3}}\right)}{\mathrm{2}×\mathrm{2000}}\right] \\ $$$$\:\:\:\:\:\:\:\approx\mathrm{0}.\mathrm{068}\:{m}\:<\mathrm{0}.\mathrm{10}\:{m}\:\checkmark \\ $$

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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