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Question Number 25943 by tawa tawa last updated on 16/Dec/17

Answered by ajfour last updated on 16/Dec/17

friction f =μN =μm_P  g              =0.5×10×10=50 newtons  m_Q g−T=m_Q a        ...(i)  T−f =m_P  a            ...(ii)  adding (i) and (ii):  m_Q g−f=(m_P +m_Q )a  or    150−50=25a  common acc. a =4m/s^2  .  T=f+m_P  a     =50+10×4 =90 newtons .  its m_Q  g  that causes P and Q  to accelerate .

$${friction}\:{f}\:=\mu{N}\:=\mu{m}_{{P}} \:{g} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{0}.\mathrm{5}×\mathrm{10}×\mathrm{10}=\mathrm{50}\:{newtons} \\ $$$${m}_{{Q}} {g}−{T}={m}_{{Q}} {a}\:\:\:\:\:\:\:\:...\left({i}\right) \\ $$$${T}−{f}\:={m}_{{P}} \:{a}\:\:\:\:\:\:\:\:\:\:\:\:...\left({ii}\right) \\ $$$${adding}\:\left({i}\right)\:{and}\:\left({ii}\right): \\ $$$${m}_{{Q}} {g}−{f}=\left({m}_{{P}} +{m}_{{Q}} \right){a} \\ $$$${or}\:\:\:\:\mathrm{150}−\mathrm{50}=\mathrm{25}{a} \\ $$$${common}\:{acc}.\:\boldsymbol{{a}}\:=\mathrm{4}{m}/{s}^{\mathrm{2}} \:. \\ $$$${T}={f}+{m}_{{P}} \:{a} \\ $$$$\:\:\:=\mathrm{50}+\mathrm{10}×\mathrm{4}\:=\mathrm{90}\:{newtons}\:. \\ $$$${its}\:{m}_{{Q}} \:{g}\:\:{that}\:{causes}\:{P}\:{and}\:{Q} \\ $$$${to}\:{accelerate}\:. \\ $$

Commented by tawa tawa last updated on 16/Dec/17

Wow, God bless you sir.

$$\mathrm{Wow},\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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