Conservation of energy problem with an incline.?

Cortez Badolato: Here's what we know. When that cart reaches max height, H, above ground level, it's kinetic energy KE = 1/2 mV^2 will be zero, V = 0 is by definition the point where max height is reached.So, from the COE law, TE = KE = 1/2 mV^2 = pe + qe = mgH + kNS = TE. Note qe is friction heat lost.mV^2 = 2mgS sin(theta) + 2kmg cos(theta)S = S2mg(sin(theta) + k cos(theta)) and S = V^2/(2g(sin(theta) + k cos(theta)) = 16/(2*9.8*(sin(radians(20)) + .2*cos(radians(20)))) = 1.540358863 = 1.54 m up the ramp. ANS....Show more

Iris Shawcroft: I think the reason you're lost is that E2 DOES NOT EQUAL E1 in this problem!Mechanical energy (that is, kinetic plus potential energy) IS NOT CONSERVED when there's friction involved. The correct conservation equation in that case is:E1 = E2 + (energy lost to friction)The energy lost to friction equals the negative work done by friction; that is: the force of friction multiplied by the distance traveled:Force of friction = (no! rmal force)×μ = mg(cosθ)μDistance traveled = d (value not yet known)energy lost to friction = mg(cosθ)μdNow let's combine that with E1, E2:E1 = (KE + PE) at bottom = ½mv²E2 = (KE + PE) at top = mghSo the conservation equation now looks like:½mv² = mgh + mg(cosθ)μdWe can eliminate one variable by noticing that h = (d)sinθ (draw a diagram to see):½mv² = mg(d)sinθ + mg(cosθ)μdFinally, just solve that for "d". (All the other variables' values are given in the problem.)...Show more