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A block of mass 10 kg, moving in x-direction with a constant speed of `10ms^(-1)`, is subjected to a retarding force `F=0.1xxJ//m` during its travel from x=20 m to 30 m. Its final KE will beA. 475JB. 450JC. 275JD. 250J |
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Answer» Correct Answer - a (a) From work-energy theorem, ` "Work done= Change in KE"` `implies W=K_(f)-K_(i)` `implies K_(f)=W+K_(i)=int_(x_(1))^(x_(2)) Fxdx+(1)/(2)mv^(2)` `=int_(20)^(30)-0.1.x.dx+1/(2)xx10xx10^(2)` `=0.1[(x^(2))/(2)]_(30)^(20)+500` `=-0.05[30^(2)-20^(2)]+500` `=-0.05[900-400]+500` `implies K_(f)=-25+500=475J` |
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