Compute the kinetic energy loss. That equals the heat generated by friction.
KE loss = Q = (M/2)[7.0^2 - 4.7^2]
= 538 J = 128.6 calories
Divide that by the heat of fusion (80 cal/g) for the mass that melts (in grams).
A 40-kg block of ice at 0°C is sliding on a horizontal surface. The initial speed of the ice is 7.0 m/s and the final speed is 4.7 m/s. Assume that the part of the block that melts has a very small mass and that all the heat generated by kinetic friction goes into the block of ice, and determine the mass of ice that melts into water at 0°C.
2 answers
Q=-KE= 1/2(M){Vf^2-Vi^2}=.5*40(4.7^2-7^2)=538.2J
Mass melted=538.2/33.5*10^4=1.6*10^-3kg
Mass melted=538.2/33.5*10^4=1.6*10^-3kg
1 answer