Conservation of energy, PE & KE problem. help?? a block is sent sliding down a frictionless ramp. Its speeds at points A and B are 2.00 m/s and 2.60 m/s, respectively. Next, it is again sent sliding down the ramp, but this time its speed at point A is 4.00 m/s. What then is its speed at point B? I understand the first step is to calculate the KE KE at pt A: KEa = (1/2) m(2)^2 KEa= 2m* calculating the KE at pt B: KEb = (1/2)mv^2 KEb = 3.38m This part from the solution guide that I'm confused about: "the difference between the KE at pt B and at pt A is the gravitational potential energy, therefore Eg = KEb - KEa Eg = 3.38 - 2m Eg = 1.38m (m is the mass)" I don't understand this part! So PE = KEa - KEb?? Why is that? And then (this part I also do not understand) KEa2 = 1/2m(4.00)^(2) = 8m To get the answer it would be 8m+1.38m=9.38m 1/2mv^(2) = 9.38m V= 4.33 m/s I am very confused. Can someone please explain to one to me. Thanks !