Question

A 2.00 kg box slides down a rough incline plane from a height h of 1.03 m. The box had a speed of 2.33 m/s at the top and a speed of 1.90 m/s at the bottom. Calculate the mechanical energy lost due to friction (as heat, etc.).

Answer 1

initial potential energy of the block = PEi = m*g*h = 2*9.8*1.03 = 20.19 J

initial kinetic energy of the ball = KEi = 0.5*m*vi^2 = 0.5*2*2.33^2 = 5.43 J

total iniital energy of the system = TEi = KEi + PEi

TEi = 25.62 J

at the bottom potential energy PEf = 0

kinetic energy = KEf = 0.5*m*vf^2 = 0.5*2*1.9^2 = 3.61 J

total energy at the bottom TEf = PEf + KEf = 3.61 J

from work energy theorem total work dione = change in energy

W =TEi - TEf

W = 25.62 -3.61 = 22.01 J   <---------answer