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Question-94158




Question Number 94158 by seedhamaieng@gmail.com last updated on 17/May/20
Commented by prakash jain last updated on 17/May/20
mass m  acceleration a=−(R/m)       −ve acclerarion as mention in        question  using relation v^2 −u^2 =2as  v^2 −u^2 =(2)(−(R/m))s  ⇒(1/2)m(u^2 −v^2 )=Rs
$$\mathrm{mass}\:{m} \\ $$$$\mathrm{acceleration}\:{a}=−\frac{{R}}{{m}} \\ $$$$\:\:\:\:\:−{ve}\:{acclerarion}\:{as}\:{mention}\:{in} \\ $$$$\:\:\:\:\:\:{question} \\ $$$$\mathrm{using}\:\mathrm{relation}\:{v}^{\mathrm{2}} −{u}^{\mathrm{2}} =\mathrm{2}{as} \\ $$$${v}^{\mathrm{2}} −{u}^{\mathrm{2}} =\left(\mathrm{2}\right)\left(−\frac{{R}}{{m}}\right){s} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{2}}{m}\left({u}^{\mathrm{2}} −{v}^{\mathrm{2}} \right)={Rs} \\ $$
Commented by i jagooll last updated on 17/May/20
very   small...
$$\mathrm{very}\:\:\:\mathrm{small}… \\ $$
Commented by seedhamaieng@gmail.com last updated on 17/May/20
thanks sir

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