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Chapter 10 Energy. Kinetic Energy and Gravitational Potential Energy We can rewrite. Kinetic Energy:. We define K = ½ mV 2 Unit of kinetic energy (Kg m 2 /s 2 ) Joule Ex. For a mass 0.5 Kg, V = 4m/s, K = 4J Gravitational potential energy: • We define U g =mgy
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Chapter 10 Energy Kinetic Energy and Gravitational Potential Energy We can rewrite
Kinetic Energy: • We define K = ½ mV2 • Unit of kinetic energy (Kg m2/s2) Joule Ex. For a mass 0.5 Kg, V = 4m/s, K = 4J Gravitational potential energy: • We define Ug=mgy • Unit of potential energy: Joule • Kinetic energy never be negative. • gravitational potential energy depends on the position.
Stop to think 10.1 P273Stop to think 10.2 P275Stop to think 10.3 P278Stop to think 10.4 P280Stop to think 10.5 P284Stop to think 10.6 P292 • Example 10.3 P275 • Example 10.4 P276 • Example 10.8 P283 • Example 10.9 P286 • Example 10.10 P291
Elastic Collisions • Perfectly elastic collision conserves both momentum and kinetic energy • Momentum conservation • Kinetic energy conservation
Perfectly elastic collision with ball 2 initially at rest • Solve above equations, the solution is Question: If m1=m2, Vi1=V, m2 initially at rest, after collision what is Vf2, Vf1
Gravitational potential energy • Gravitational potential energy: mgh Y or (h) depends on where you choose to put the origin of your coordinate system. But potential energy change ΔU is independent of the coordinate system
Quick think • A small child slides down the four frictionless slides A-D. Each has the same height. Rank in order. From largest to smallest her speeds VA to VD at the bottom. VA = VB =VC =VD
Ex. 10.4 A ballistic pendulumA 10 g bullet is fired into a 1200g wood block hanging from a 150-cm-long string. The bullet embeds itself into the block, and block then swings out to an angle of 40o. What was the speed of the bullet? The momentum conservation equation Pi = Pf applied to the inelastic collision Then turning our attention to the swing The energy equation Kf + Ugf = Ki + Ugi We define y1 = 0 Get:
Three identical balls are thrown from the top of a building, all with the same initial speed the first is thrown horizontally, the second at some angle above the horizontal and third at some angle below the horizontal.Neglecting the air resistance, rank the speeds of the balls at the instant each hits the ground • . • Answer: All the three balls have the same speed at the moment they hit the ground. • Since neglect the air resistance, total mechanical energies for each ball should be conserved. • Ki+Ui= 1/2mvi2 +mgh • Kf + Uf = 1/2mvf2 • Ki+Ui = Kf +Uf • Vf2= 2gh + vi2 does not matter the angle. • Three balls take different times to reach the ground
Restoring forces and Hooke’s Law • Displacement from equilibrium ∆s • Hooke’s Law • K is called the spring constant. It is spring character, unit:N/m
Conservation of mechanical energy • Mechanical energy E (mech) = K + U • If there is no friction or other losses of mechanical energy then ΔE(mech)=0 This is the law of conservation of mechanical energy
Energy Diagrams Particle in free fall • Energy Diagram is used to visualize motion.
More general energy diagram How to get kinetic energy from this diagram?