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))

A block of ice of mass 50kg is sliding on a horizontal plane. It start

1.	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» 2.86 G 3.86 g 0.86 g 1.86 g Solution :d) Given, m = 50kg, v=`5ms^(-1)`, L = 80 `calg^(-1)` J=4.2J `cal^(-1)` and m=? Heat lost i.e., `Q_(1)` = workd DONE, K = Kinetic energy i.e., `1/2mv^(2)= 1/2 xx50 xx5^(2)`=625 J and heat GAINED i.e., `Q_(2)=m^(')xxL = 80 xxm^(')xx4.2` `THEREFORE` `Q_(1)=Q_(2)` 625 = `m^(') xx 8 xx 42 rArr m^(')=1.86`g