Physics exercise, psychology homework help

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nygnjvy89

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I need step by step answer for the attached exercise.

I need solution to the same exercise but using a frictional method on both sides A to B B to C.

I also attached one more example to review if needed..

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Note: When you have friction, don't write TEATE lon Conservative work) friction part A -> B: use non-friction part В с Wnc (non TE (Total energy) 3.0 B C : use C M=0.2 30 l I K-400 has friction WIN O f 여 a my AB - fid= AKE + APE - Mn. 3.0= ( 2 m 3 - 2 1 ) + ( (mshing - ngh) -(0.2) (2g cos 30) (3,0) = 4121 18+ [=2913.0g sin 30) - 10,18 VB + 2881 ing caso [img sino fe oppo to the dir of motion 0 کر ^ 2 X V 2 B 29.4 29.4 10 18 2 있 & 1916 1417 Syson VB = 4.6 Then: TEB = TEC En voi i x = y + ingh & + KX ? 2 m = -2 K x Ź (21(426) 400) YE I 2 2 2 19.36 - 200 х? С C ya 19.36 * 19:56 - 0.31 200 A V; = 0 2kg 3m M=0.2 3.0 sin 30° 38 B C yed A>B: Te method. 1 total energy method B>C: Frictional method.) 2 Frichtond method AB: TEA =TER Žmv + mg h = £ mv + mghz 3 missing part N3 5.4 m/s B-C: (-fod=AKE+OPE (f = Anb/-(41).d = {mx-4 m vj + (myghe - might n = mg) -(0.2)(zg)(d) = -{(z) V
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Here is the answer in word and pdf file.

Exercise (1):

(1) Using TE method from A to B:
𝑻𝑬𝑨 = 𝑻𝑬𝑩
𝟏
𝟏
𝒎𝒗𝟐𝑨 + 𝒎𝒈𝒉𝑨 = 𝒎𝒗𝟐𝑩 + 𝒎𝒈𝒉𝑩
𝟐
𝟐
𝟏
𝟏
× 𝟐 × 𝟎𝟐 + 𝟐 × 𝟗. 𝟖 × 𝟑𝒔𝒊𝒏(𝟑𝟎) = × 𝟐 × 𝒗𝟐𝑩 + 𝟐𝒈 × 𝟎
𝟐
𝟐
𝟐 × 𝟗. 𝟖 × 𝟑𝒔𝒊𝒏(𝟑𝟎) = 𝒗𝟐𝑩
𝒗𝟐𝑩 = 𝟐𝟗. 𝟒
𝒗𝑩 = √𝟐𝟗. 𝟒 = 𝟓. 𝟒 𝒎/𝒔
(2) Using friction method from B to C:
−𝒇 × 𝒅 = ∆𝑲. 𝑬 + ∆𝑷. 𝑬
𝑭=𝝁×𝒏
𝒏 = 𝒎𝒈

𝟏

𝟏

𝟐

𝟐
𝟏

−𝝁 × 𝒏 × 𝒅 = ( 𝒎𝒗𝟐𝑪 − 𝒎𝒗𝟐𝑩 ) + (𝒎𝒈𝒉𝑪 − 𝒎𝒈𝒉𝑩 )
𝟏

−𝝁 × 𝒎𝒈 × 𝒅 = ( 𝒎 × 𝟎𝟐 − 𝒎𝒗𝟐𝑩 ) + (�...


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