Conservation of Energy If only conservative forces are present, the total energy (sum of potential and kinetic energies ) of a system is conserved E=K+U △E=△K+△U =W+△U using△K=W W+(-W=0 using△U=-W E=K+ U is constant∥! Active Figure Both K and U can change but E=K+Uremains constant Physics 121: Lecture 14, Pg 6
Physics 121: Lecture 14, Pg 6 Conservation of Energy If only conservative forces are present, the total energy (sum of potential and kinetic energies) of a system is conserved. E = K + U E = K + U = W + U = W + (-W) = 0 using K = W using U = -W Both K and U can change, but E = K + U remains constant. E = K + U is constant !!! Active Figure
Problem: Hotwheel a toy car slides on the frictionless track shown below It starts at rest, drops a distance d, moves horizontally at speed V,, rises a distance h, and ends up moving horizontally with speed v, Find v, and v2 h Physics 121: Lecture 14, Pg 7
Physics 121: Lecture 14, Pg 7 Problem: Hotwheel. A toy car slides on the frictionless track shown below. It starts at rest, drops a distance d, moves horizontally at speed v1 , rises a distance h, and ends up moving horizontally with speed v2 . Find v1 and v2 . h d v1 v2
Problem: Hotwheel Energy is conserved, So AE=0 △KE=-△PE Moving down a distance d,△PE=mg,△KE=1mv2 Solving for the speed 2gd h Physics 121: Lecture 14, Pg 8
Physics 121: Lecture 14, Pg 8 Problem: Hotwheel... Energy is conserved, so E = 0 KE = - PE Moving down a distance d, PE = -mgd, KE = 1 /2mv1 2 Solving for the speed: h d v1 v1 = 2gd
Problem: Hotwheel At the end we are a distance d-h below our starting point △PE=-mg(d-h),△KE=2mv2 Solving for the speed 2g(d-h d-h h Physics 121: Lecture 14, Pg 9
Physics 121: Lecture 14, Pg 9 Problem: Hotwheel... At the end, we are a distance d-h below our starting point. PE = -mg(d-h), KE = 1 /2mv2 2 Solving for the speed: h d v2 v g(d h) 2 = 2 − d-h