1. The problem statement, all variables and given/known data A block of ice with mass 2.00 kg slides 0.750 m down an inclined plane that slopes downward at an angle of 36.9° below the horizontal. If the block of ice starts from rest, what is its final speed? You can ignore friction. This problem is found in the University Physics With Modern Physics 13th Edition Book. It is problem number 6.28 in Chapter 6. 2. Relevant equations Wtot = Ʃ F times d Wtot = 1/2mv^2 (final K2) - 1/2 mv^2 (initial K1) 3. The attempt at a solution Hello guys! I hope everyone is well! There is something I'm not really quite understanding about this particular problem. Heres my attempt at the solution: First I found the Work Total done. I went ahead and found the parallel component of the ice blocks weight to the slope and multiplied it by the distance of 0.750m down the slope. This gave me a Work Total of 11.75J I then used the formula Wtot = 1/2mv^2 (final) - 1/2mv^2 (initial) with K1 being 0 since the ice block starts from rest. 11.75J = 1/2(2kg)v^2 - 0 divided 11.75J by 1kg to give me 11.75 m^2/s^2 took sqrt of 11.75 m^2/s^2 V=√11.75 m^2/s^2 = 3.42 m/s The answer in the book is 2.97m/s. Im not sure where I went wrong exactly, and Im having a hard time trying to figure out the correct solution. I would love some help! Thanks!