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A block of ice of mass 50kg is sliding on a horizontal plane. It starts with speed 5ms^(-1) and stop 3 after moving through some distance. The mass of ice that has melted due to friction between the block and the suface is (assuming that no energy is lost nad the surface is (assuming that no energy is lost and latent heat of fustion of ice is 80 calg^(-1))

Answer» <html><body><p>2.86 <a href="https://interviewquestions.tuteehub.com/tag/g-1003017" style="font-weight:bold;" target="_blank" title="Click to know more about G">G</a><br/>3.86 g<br/>0.86 g<br/>1.86 g</p>Solution :d) Given, m = 50kg, v=`5ms^(-1)`, L = 80 `calg^(-1)` <br/> <a href="https://interviewquestions.tuteehub.com/tag/j-520843" style="font-weight:bold;" target="_blank" title="Click to know more about J">J</a>=4.2J `cal^(-1)` and m=? <br/> Heat lost i.e., `Q_(1)` = workd <a href="https://interviewquestions.tuteehub.com/tag/done-2591742" style="font-weight:bold;" target="_blank" title="Click to know more about DONE">DONE</a>, K = Kinetic energy <br/> i.e., `1/2mv^(2)= 1/2 xx50 xx5^(2)`=625 J <br/> and heat <a href="https://interviewquestions.tuteehub.com/tag/gained-2665852" style="font-weight:bold;" target="_blank" title="Click to know more about GAINED">GAINED</a> i.e., `Q_(2)=m^(')xxL = 80 xxm^(')xx4.2` <br/> `<a href="https://interviewquestions.tuteehub.com/tag/therefore-706901" style="font-weight:bold;" target="_blank" title="Click to know more about THEREFORE">THEREFORE</a>` `Q_(1)=Q_(2)` <br/> 625 = `m^(') xx 8 xx 42 rArr m^(')=1.86`g</body></html>


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