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PHYSICS 231 Lecture 11: How much energy goes into problems?

PHYSICS 231 Lecture 11: How much energy goes into problems?. Contact me! Thompson R. (003) Conti M. (001) Zaran E. (002) Whymer K. (002) Aggarwal S. (?) Chirio A. (001) Webb A. (002) Leboeuf J. (002) McKenzie A. (002). Remco Zegers Walk-in hour: Thursday 11:30-13:30 am Helproom.

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PHYSICS 231 Lecture 11: How much energy goes into problems?

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  1. PHYSICS 231Lecture 11: How much energy goes into problems? Contact me! Thompson R. (003) Conti M. (001) Zaran E. (002) Whymer K. (002) Aggarwal S. (?) Chirio A. (001) Webb A. (002) Leboeuf J. (002) McKenzie A. (002) Remco Zegers Walk-in hour: Thursday 11:30-13:30 am Helproom PHY 231

  2. Previous lecture • Work: W=Fcos()x Energy transfer • Power: P=W/t Rate of energy transfer • Potential energy (PE) Energy associated with position. • Gravitational PE: mgh Energy associated with position in grav. field. • Kinetic energy KE: ½mv2 Energy associated with motion • Conservative force: Work done does not depend on path • Non-conservative force: Work done does depend on path • Mechanical energy ME: ME=KE+PE • Conserved if only conservative forces are present KEi+PEi=KEf+PEf • Not conserved in the presence of non-conservative forces (KEi+PEi)-(KEf+PEf)=Wnc PHY 231

  3. Work and energy WORK POTENTIAL ENERGY KINETIC ENERGY PHY 231

  4. Mechanical Energy Mechanical energy Gravitational Potential Energy (mgh) Kinetic Energy ½mv2 PHY 231

  5. Overview Conservation of mechanical energy Newton’s second Law F=ma Wnc=0 Closed system Work W=(Fcos)x Work-energy Theorem Wnc=Ef-Ei Equations of kinematics X(t)=X(0)+V(0)t+½at2 V(t)=V(0)+at PHY 231

  6. (PE1+PE2+KE1+KE2)=constant At time of release: PE1=m1gh1=5.00*9.81*4.00 =196. J PE2=m2gh2=3.00*9.81*0.00 =0.00 J KE1=0.5*m1*v2=0.5*5.00*(0.)2 =0.00 J KE2=0.5*m1*v2=0.5*3.00*(0.)2 =0.00 J Total =196. J At time of passing: PE1=m1gh1=5.00*9.81*2.00 =98.0 J PE2=m2gh2=3.00*9.81*0.00 =58.8 J KE1=0.5*m1*v2=0.5*5.00*(v)2 =2.5v2 J KE2=0.5*m1*v2=0.5*3.00*(v)2 =1.5v2 J Total =156.8+4.0v2 196=156.8+4.0v2 so v=3.13 m/s Conservation of mechanical energy A) what is the speed of m1 and m2 when they pass each other? PHY 231

  7. W=Fx=m1g2.00+m2g(-2.00)=39.2 J Pestart- Pepassing=(196.-98.-58.8)= 39.2 J The work done by Fg is the same as the change in potential energy work How much work is done by the gravitational force when the masses pass each other? PHY 231

  8. (PE+ KE)start-(PE+KE)passing=Wnc Wnc=Ffrictionx=5.00*2.00=10.0 J (196.)-(156.8+KE)=10 J KE=29.2 J=0.5*(m1+m2)v2=4v2 v=2.7 m/s Friction (non-conservative) The pulley is not completely frictionless. The friction force equals 5 N. What is the speed of the objects when they pass? PHY 231

  9. A spring k: spring constant (N/m) Fs=-kx +x Fs(x=0)=0 N Fs(x=-a)=ka Fs=(0+ka)/2=ka/2 Ws=Fsx=(ka/2)*(a)=ka2/2 The energy stored in a spring depends on the location of the endpoint: elastic potential energy. PHY 231

  10. PINBALL The ball-launcher spring has a constant k=120 N/m. A player pulls the handle 0.05 m. The mass of the ball is 0.1 kg. What is the launching speed? (PEgravity+PEspring+KEball)pull=(PEgravity+PEspring+KEball)launch mghpull+½kxpull2+½mvpull2 = mghlaunch+½kxlaunch2+½mvlaunch2 0.1*9.81*0+½120(0.05)2+½0.1(0)2= 0.1*9.81*(0.05*sin(10o))+½120*(0)2+½0.1vpull2 0.15=8.5E-03+0.05v2 v=1.7 m/s PHY 231

  11. Ball on a track A h end B h end In which case has the ball the highest velocity at the end? A) Case A B) Case B C) Same speed In which case does it take the longest time to get to the end? A) Case A B) Case B C) Same time PHY 231

  12. With friction KE PE TME NC KE PE TME NC KE PE TME NC KE PE TME NC KE PE TME NC Race track PHY 231

  13. A swing If relieved from rest, what is the velocity of the ball at the lowest point? 30o L=5m (PE+KE)=constant PErelease=mgh (h=5-5cos(30o)) =6.57m J KErelease=0 PEbottom=0 KEbottom=½mv2 ½mv2=6.57m so v=3.6 m/s h PHY 231

  14. A running person While running, a person dissipates about 0.60 J of mechanical energy per step per kg of body mass. If a 60 kg person develops a power of 70 Watt during a race, how fast is she running (1 step=1.5 m long) What is the force the person exerts on the road? W=Fx P=W/t=Fv Work per step: 0.60 J/kg * 60 kg=36 J Work during race: 36*(racelength(L)/steplength)=24L Power= W/t=24L/t=24vaverage=70 so vaverage=2.9 m/s F=P/v so F=24 N PHY 231

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