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PHYSICS 231 Lecture 10: Too much work!

PHYSICS 231 Lecture 10: Too much work!. Remco Zegers. 60N. 50 kg (top view). 80N. Two persons are dragging a box over a floor. Assuming there is no friction, what is the acceleration of the crate? 1.2 m/s 2 1.6 m/s 2 2.0 m/s 2 2.8 m/s 2 no acceleration whatsoever. WORK.

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PHYSICS 231 Lecture 10: Too much work!

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  1. PHYSICS 231Lecture 10: Too much work! Remco Zegers PHY 231

  2. 60N 50 kg (top view) 80N • Two persons are dragging a box over a floor. Assuming there • is no friction, what is the acceleration of the crate? • 1.2 m/s2 • 1.6 m/s2 • 2.0 m/s2 • 2.8 m/s2 • no acceleration whatsoever PHY 231

  3. WORK Work: ‘Transfer of energy’ Quantitatively: The work W done by a constant force on an object is the product of the force along the direction of displacement and the magnitude of displacement. W=(Fcos)x Units: =Nm=Joule PHY 231

  4. An example n n T =45o T fk x Fg Fg The work done by the person on the suitcase: W=(Tcos45o)x The work done by Fg on the suitcase: W=(Fgcos270o)x=0 The work done by n on the suitcase: W=(Fgcos90o)x=0 The work done by friction on the suitcase: W=(fkcos180o)x=-uknx The work done by the suitcase on the person: W=(Tcos225o)x opposite PHY 231

  5. Area=A=(Fcos)x W=(A) Fcos Fcos The work done is the same as the area under the graph of Fcos versus x Non-constant force W=(Fcos)x: what if Fcos not constant while covering x? Example: what if  changes while dragging the suitcase? PHY 231

  6. Power: The rate of energy transfer Work (the amount of energy transfer) is independent of time. W=(Fcos)x … no time in here! To measure how fast we transfer the energy we define: Power(P)=W/t (J/s=Watt) (think about horsepower etc). P =(Fcos)x/t=(Fcos)vaverage PHY 231

  7. Example A toy-rocket of 5.0 kg, after the initial acceleration stage, travels 100 m in 2 seconds. What is the work done by the engine? What is the power of the engine? W=(Fcos)h=mrocketg h=4905 J (Force by engine must balance gravity!) P=W/t=4905/2=2453 W (=3.3 horsepower) or P=(Fcos)v=mgv=5.0 9.81 100/2=2453 W h=100m PHY 231

  8. Potential Energy Potential energy (PE): energy associated with the position of an object within some system. Gravitational potential energy: Consider the work done by the gravity in case of the rocket: Wgravity=Fg cos(180o)h=-mgh=-(mghf-mghi)=mghi-mghf =PEi-PEf The ‘system’ is the gravitational field of the earth. PE=mgh Since we are usually interested in the change in gravitational potential energy, we can choose the ground level (h=0) in a convenient way. PHY 231

  9. h(t)=h(0)+v0t+0.5at2 100=0.5a22 so a=50 m/s2 V(t)=V(0)+at V(2)=0+50*2=100 m/s Force by engine=(50+9.81)m=59.81*5=299.05 N (9.81 m/s2 due to balancing of gravitation) W=Fh=299.05*100=29905 J Another rocket A toy rocket (5kg) is launched from rest and reaches a height of 100 m within 2 seconds. What is the work done by the engine during acceleration? h=100m Change in potential energy: PEf-PEi=mgh(2s)-mgh(0)=4905-0=4905 J Where did all the work (29905-4905=25000 J)go? Into the acceleration: energy of motion (kinetic energy) PHY 231

  10. x=100m Kinetic energyConsider object that changes speed only V=0 t=2s a) W=Fx=(ma)x … used Newton’s second law b) v=v0+at so t=(v-v0)/a c) x=x0+v0t+0.5at2 so x-x0=x=v0t+0.5at2 Combine b) & c) d) ax=(v2-v02)/2 Combine a) & d) W=½m(v2-v02) Kinetic energy: KE=½mv2 When work is done on an object and the only change is its speed: The work done is equal to the change in KE: W=KEfinal-KEinitial Rocket: W=½5(1002-02) =25000 J!! That was missing! PHY 231

  11. Conservation of mechanical energy Mechanical energy = potential energy + kinetic energy In a closed system, mechanical energy is conserved* V=100 ME=mgh+½mv2=constant 5 kg What about the accelerating rocket? At launch: ME=5*9.81*0+½5*02=0 At 100 m height: ME=5*9.81*100+½5*1002=29905 h=100m We did not consider a closed system! (Fuel burning) * There is an additional condition, see next lecture V=0 PHY 231

  12. question energy A time B energy In the absence of friction, which energy-time diagram is correct? time C energy potential energy total energy kinetic energy time PHY 231

  13. At ground: ah=(v2-v0,ver2)/2 so v=26.2 m/s V=(Vhor2+Vver2)=(12.72+26.22)=29.1 ME=mgh+½mv2 =0.2*9.81*0+½*0.2*29.12=84.7 J ME is conserved!! Example of closed system A snowball is launched horizontally from the top of a building at v=12.7 m/s. The building is 35m high. The mass is 0.2 kg. Is mechanical energy conserved? V0=12.7 m/s At launch: ME=mgh+½mv2 =0.2*9.81*35+½*0.2*12.72=84.7 J h=35m d=34m PHY 231

  14. mgh=1/2mv2 v=(2gh) h end Consider the above rollercoaster (closed system). The cart starts at a height h. What is its velocity v at the end? Hint: consider the mechanical energy in the beginning and the end. a) v=2gh b) v=gttrip c) v=(2gh) d) v=(2gh/m) e) v=0 m/s kinetic energy: 0.5mv2 potential energy: mgh PHY 231

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