Physics, asked by dewanganpatwari, 10 months ago

In the figure find acceleration

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Answers

Answered by Deepsonloit

Answer:

there is no acceleration.

Explanation:

Because acceleration means the change in the velocity per unit time.

Answered by FlowerBlush

Answer :-

acceleration = (m+M+Mμmg)i^+(m+M+Mμmg)j^

Solution :-

Let the string has T tension, and accelerators

μ= friction coefficient

μ= friction coefficientBy force equation:

μ= friction coefficientBy force equation:T=Ma

μ= friction coefficientBy force equation:T=Ma x

μ= friction coefficientBy force equation:T=Ma x -----eq.1

μ= friction coefficientBy force equation:T=Ma x -----eq.1mg−T−Tμ=ma

μ= friction coefficientBy force equation:T=Ma x -----eq.1mg−T−Tμ=ma y

μ= friction coefficientBy force equation:T=Ma x -----eq.1mg−T−Tμ=ma y -----eq.2

μ= friction coefficientBy force equation:T=Ma x -----eq.1mg−T−Tμ=ma y -----eq.2equation of the string:

μ= friction coefficientBy force equation:T=Ma x -----eq.1mg−T−Tμ=ma y -----eq.2equation of the string:y - x =

-----eq.3

-----eq.3From eq.1 and eq.2

-----eq.3From eq.1 and eq.2mg−Ma

-----eq.3From eq.1 and eq.2mg−Ma x

-----eq.3From eq.1 and eq.2mg−Ma x −Ma

-----eq.3From eq.1 and eq.2mg−Ma x −Ma x

-----eq.3From eq.1 and eq.2mg−Ma x −Ma x μ=ma

-----eq.3From eq.1 and eq.2mg−Ma x −Ma x μ=ma y

-----eq.3From eq.1 and eq.2mg−Ma x −Ma x μ=ma y Form eq.3

-----eq.3From eq.1 and eq.2mg−Ma x −Ma x μ=ma y Form eq.3mg−Ma

-----eq.3From eq.1 and eq.2mg−Ma x −Ma x μ=ma y Form eq.3mg−Ma x

-----eq.3From eq.1 and eq.2mg−Ma x −Ma x μ=ma y Form eq.3mg−Ma x −Ma

-----eq.3From eq.1 and eq.2mg−Ma x −Ma x μ=ma y Form eq.3mg−Ma x −Ma x

-----eq.3From eq.1 and eq.2mg−Ma x −Ma x μ=ma y Form eq.3mg−Ma x −Ma x μ=ma

-----eq.3From eq.1 and eq.2mg−Ma x −Ma x μ=ma y Form eq.3mg−Ma x −Ma x μ=ma x

-----eq.3From eq.1 and eq.2mg−Ma x −Ma x μ=ma y Form eq.3mg−Ma x −Ma x μ=ma x ma

-----eq.3From eq.1 and eq.2mg−Ma x −Ma x μ=ma y Form eq.3mg−Ma x −Ma x μ=ma x ma x

-----eq.3From eq.1 and eq.2mg−Ma x −Ma x μ=ma y Form eq.3mg−Ma x −Ma x μ=ma x ma x +Ma

-----eq.3From eq.1 and eq.2mg−Ma x −Ma x μ=ma y Form eq.3mg−Ma x −Ma x μ=ma x ma x +Ma x

-----eq.3From eq.1 and eq.2mg−Ma x −Ma x μ=ma y Form eq.3mg−Ma x −Ma x μ=ma x ma x +Ma x +Ma

-----eq.3From eq.1 and eq.2mg−Ma x −Ma x μ=ma y Form eq.3mg−Ma x −Ma x μ=ma x ma x +Ma x +Ma x

-----eq.3From eq.1 and eq.2mg−Ma x −Ma x μ=ma y Form eq.3mg−Ma x −Ma x μ=ma x ma x +Ma x +Ma x μ=mg

-----eq.3From eq.1 and eq.2mg−Ma x −Ma x μ=ma y Form eq.3mg−Ma x −Ma x μ=ma x ma x +Ma x +Ma x μ=mga

-----eq.3From eq.1 and eq.2mg−Ma x −Ma x μ=ma y Form eq.3mg−Ma x −Ma x μ=ma x ma x +Ma x +Ma x μ=mga x

-----eq.3From eq.1 and eq.2mg−Ma x −Ma x μ=ma y Form eq.3mg−Ma x −Ma x μ=ma x ma x +Ma x +Ma x μ=mga x =

-----eq.3From eq.1 and eq.2mg−Ma x −Ma x μ=ma y Form eq.3mg−Ma x −Ma x μ=ma x ma x +Ma x +Ma x μ=mga x = m+M+Mμ

-----eq.3From eq.1 and eq.2mg−Ma x −Ma x μ=ma y Form eq.3mg−Ma x −Ma x μ=ma x ma x +Ma x +Ma x μ=mga x = m+M+Mμmg

-----eq.3From eq.1 and eq.2mg−Ma x −Ma x μ=ma y Form eq.3mg−Ma x −Ma x μ=ma x ma x +Ma x +Ma x μ=mga x = m+M+Mμmg acceleration =

-----eq.3From eq.1 and eq.2mg−Ma x −Ma x μ=ma y Form eq.3mg−Ma x −Ma x μ=ma x ma x +Ma x +Ma x μ=mga x = m+M+Mμmg acceleration = m+M+Mμ

-----eq.3From eq.1 and eq.2mg−Ma x −Ma x μ=ma y Form eq.3mg−Ma x −Ma x μ=ma x ma x +Ma x +Ma x μ=mga x = m+M+Mμmg acceleration = m+M+Mμmg

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