= KE + U. Only conservative forces (gravity & spring) cause energy transfer (work)

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2 Mechanical energy: h = KE + U Only conservative forces (gravity & spring) cause energy transfer (work) W net = ΔKE Sec. 7-3 W conservative = ΔU Sec. 8-1 Isolated system: Assuming only internal forces (no external forces yet - Sec. 8.6) No external force from outside causes energy change inside Δh = 0 = Δ KE + U ( ) ΔKE = ΔU

3 F(x) = du(x) dx F grav (y) = du grav (y) dy F spring (x) = du spring (x) dx ( = mgy ) dy = = mg 1 ( kx 2 2 ) = kx dx Equilibrium Points Turning point h = 3 J h = 5 J h = 4 J h = 1 J Equilibrium positions: where slope of U(x) curve is zero [i.e. F(x) = 0 ; NO FORCE ] -> Neutral vs Unstable vs Stable Equilibrium

4 Problem #40: The figure shows a potential energy curve U vs position. If U A =9.00J, U C =20.J, U D =24.0J. The particle is released the point on U at the point between 1 and 16.0J 3 m with a kinetic energy of 4J and U B =12. J. What is the Kinetic energy and speed at x=3.5 m and x=6.5m? Where is the turning point on the left and right. At point B: h = KE B +U B =4J+12J = 16 J At x = 3.5 m: h = KE x=3.5 +U x=3.5 =Ke x=3.5 +9J Since there is only potential energy, h is conserved Thus KE x=3.5 =h -U x=3.5 = 16 J -9 J=7 J At x = 6.5 m: h = KE x=6.5 +U x=6.5 =Ke x= This gives KE x=6.5 =h -U x=6.5 = 16 J Since h = 16 J, the particle cannot go beyond x < 2.0 m or x > 7.5 m. Therefore, x = 2.0 m and 7.5 m are turning points.

5 Problem #121: A conservative force F(x) acts on a 2.0 kg particle. The potential energy U(x) is shown in the figure. When the particle is at x=2.0 m it has a velocity of -1.5m/s. (a) find the magnitude and direction of the force at this position. (b) What is the range of the allowed motion? (c) What is the velocity at x=7.0 m? = K i + U i = 2.25J 7.5J = 5J K i = mv2 2 = 2.25J Range given by green arrows. K i = mv2 2 = J = 12J What is direction and magnitude of the force at 2, 7, 12 m?

6 Non-conservative Forces: Friction If there is no non-conservative force involved, My preference is to always use Mechanical Energy ( final) = (initial) K f + U f = K i + U i When Friction is present, it always removes energy, so E(final) is less the E(initioal) ( final) = (initial) Energy removed by work of friction K f + U f = K i + U i F f d If there is no work done by an external force Δh + ΔE Thermal = 0 h ( final) = h (initial) E th ( final) + E th (initial) K f + U f = K i + U i F f times the displacement

7 What happens if non-isolated or non-conservative forces? Friction? Motor?

8 A new way to look at friction Question #8: A block of mass m slides down the inclined plane starting with zero velocity. Region D has friction and it comes to rest after moving a distance D. Mechanical Energy is not Conserved. ( final) = (initial) ΔE th ( final) = (initial) F f x We can draw this. ( final) = (initial) ΔE th ( final) = (initial) F f x K = mv2 2

9 A new way to look at Friction Question #9: Find K and as the block moves along. h -F f distance K = mv2 2

10 Conserv Force Other Forces F gravitation F applied F spring F friction F air (drag) F tension F normal Work done by External Force Conservative Force in isolated system Δh = 0 ΔKE = ΔU If External force: (NO FRICTION) W ext,net = Δh = ΔKE + ΔU Positive if energy is transferred to system Negative if energy is transferred from system

11 Work done by External Force + Friction Example: Ext Force with Friction W ext, fric = Δh = ΔKE + ΔU W ext = F ext d W friction = f k d positive or negative always negative W ext, fric = F ext d f k d ( ) = Δh Define: ΔE therm = f k d (increase in thermal energy by sliding) W ext,net = Δh + ΔE therm = ( ΔKE + ΔU) + ΔE therm

12 More Sample Problem A 10 kg block is released from point A in the figure. The track is frictionless except for the portion between points B and C, which has a length of 6 m. The block travels down the track, hits a spring of force constant 250 N/m, and compresses the spring 0.3 m from its equilibrium position before coming to rest momentarily. Determine the coefficient of Kinetic friction between the block and the rough surface between B and C. (initial) = mgh Energy lost=f f d = mgµ k d So when it compresses the spring v=0 kx 2 2 = mgh mgµ k d

13 Problem A block slides along a track from one level to a higher level, by moving through an intermediate valley. The track is frictionless until the block reaches the higher level. There is a frictional force that stops the block in the distance d. The block s initial speed is v 0 ; the height difference is h; the coefficient of kinetic friction is µ k. What is d? What do we know? Along the part of the track which is frictionless, conservation of Mechanical Energy holds. However, at the top the friction transfers energy out of the system (ΔE therm ). Since the system is isolated the change in the total energy is zero. 0 = W ext,net = [ ΔKE + ΔU net ] + ΔE therm [( ) + ( mgh 0) ] + µ k mg 0 = 0 1 mv Solving for d d = 1 µ k mg ( ( ))d = v 2 0 2µ k g h µ k 1 ( mv 2 mgh 2 0 )

14 Problem 8-63 A particle can slide along a track as shown. The curved portions of the track are frictionless, but the flat part has a coefficient of kinetic friction of µ k = The particle is released from rest at point A, which is a height h=l/2 above the flat part of the track. Where does the particle finally stop? Step I (from A to B): h (A)=h (B) This leads: KE B =mgh B C Step II (from B to C): since there is friction force along -x direction, KE C -KE B =W fk =-μ k (mg)l thus KE C =mgh-μ k (mg)l Step III (from C to where it stops, saying D ) in this case KE C is completely transferred to potential energy mgy i.e., mgh-μ k (mg)l=mgy y = h-μ k L

15 If there is no non-conservative My force preference involved, is to always use Mechanical Energy Conservation ( final) = (initial) K f + U f = K i + U i When Friction is present, it always removes energy, so E(final) is less the E(initioal) initial ( final) = (initial) Energy removed by work of friction K f + U f = K i + U i F f d Always think about the problem and the signs!!