1. The problem statement, all variables and given/known data A 25.6 kg child pulls a 4.81kg toboggan up a hill inclined at 25.7 degrees to the horizontal. The vertical height of the hill is 27.3 m. Friction is negligible. Determine the amount of work the child must do on the toboggan to pull it at a constant velocity up the hill. 2. Relevant equations W = F * D 3. The attempt at a solution ag = gravity component of acceleration ap = acceleration applied ag - ap = 0 ag = 9.81 m/s^2 * cos(25.7 deg) = 8.8 m/s/s F = 8.8 m/s/s * 4.81 kg = 42.5 N W = F * D = 42.5 N * (27.3 m / (sin(25.8 deg)) = 2 665.82513 joules This answer is completely wrong. The book got 1.3 * 10 ^ 3 J. Not sure what I can do, I thought about setting W = Eg2 - Eg1. But I remember my teacher said that only works for kinetic energy (honestly, sometimes I think she doesn't know what shes talking about. It should work for gravitational energy too) But if I do that, W = 4.8 kg * 9.8 m/s^2 * 27.3 m I get the right answer which is 1 284.192 joules This is confusing for me. Shouldn't I have gotten the same answers for both questions? How come my answers are different? Does it have something to do with vectors? Does work only count in one direction in this case (vertical), shouldn't the work done be how I calculated it in step 1? More importantly, can I set W = change in energy, even if that energy is NOT kinetic?