1. Motion, Forces and Energy Lecture 7: Potential Energy & Conservation The name potential energy implies that the object in question has the capability of either gaining kinetic energy or doing work on some other object: Gravitational Potential Energy U g = m g h Elastic Potential EnergyU s = ½ k x 2

2. Conservation of Energy h1h1 h2h2 0 y v1v1 v2v2 Total Energy, E 1 Total Energy, E 2 The change in total energy is  E where:

3. Object in free fall A ball of mass m is dropped from a height h above the ground. We can use energy calculations to find the speed of the ball when it is either released from rest or with an initial speed v i : h y y i = h U i = mgh KE i = ½ mv i 2 y f = y U f = mgy KE f = ½ mv f 2 y=0 U g =0

4. Energy losses (non-conservative forces) Kinetic friction is an example of a non-conservative force. When a book slides across a surface which is not frictionless, it will eventually stop. But all the KE of the book is NOT transferred to internal energy of the book. We can find the speed of the mass sliding down the ramp below if we know the frictional force by considering kinetic and potential energies: v i =0 vfvf h=0.5m d=1.0m Initial energy, E i = KE i +U i = mgy i =14.7J Final energy, E f = KEf+Uf = ½ mv f 2 But E i = E f due to energy losses due to friction so energy loss  E is  E = -f k.d = -  k mgcos .d = -5.0J So ½ mv f 2 = 14.7-5.0 J And therefore v f =2.54 ms -1 30 o

5. Another similar problem: The friction coefficient between the 3 kg block and the surface is 0.4. If the system starts from rest, calculate the speed of the 5 kg ball when it has fallen a distance of 1.5 m. Mechanical energy loss is –f k.d = -  k m 1 g d (work done by friction on block) = -17.6 J Change in PE for the 5 kg mass is  U 5kg = -m 2 g d = -73.5 J Without friction, the KE gained by BOTH masses would sum to  U 5kg (73.5 J), but With friction, the KE gained = 73.5 – 17.6 = 55.9 J  KE = ½ (m 1 +m 2 )v 2 (both masses move with same velocity) So v = 3.7 ms -1. m 1 =3.0 kg m 2 =5.0 kg fkfk

6. An inclined spring A mass m starts from rest and slides a distance d down a frictionless incline. It strikes an unstressed spring (negligible mass) and slides a further distance x (compressing the spring which has a force constant, k). Find the initial separation of the mass and the end of the spring. m d  k Vertical height travelled by the mass = (d+x)sin  =h Change in PE of the mass,  U g = 0 – mgh = -mg(d+x)sin  Elastic PE of spring increases from zero (unstressed) to:  U s = ½ kx 2 So  U g +  U s = 0 So mg (d+x) sin  = ½ kx 2 giving d as: