Conservation of Energy

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Presentation transcript:

Conservation of Energy

Facts Conservation of energy When gravity is the only force which does work on a body, mechanical energy is conserved. PEstart + KEstart = PEend + KEend This is also true when an external force is applied PEstart + KEstart + Win = PEend + KEend+ Wout

Examples A toolbox of mass 3 kg, is knocked off a table which is 0.3 m above the ground. Using conservation of energy, calculate the vertical speed with which the toolbox lands on the ground. 3g 0.3 m Zero PE level As the toolbox falls the initial PE at the start is converted into KE at the end PEstart = KEend mgh = ½mv² 3 x 9.8 x 0.3 = ½ x 3 x v² v² = 5.88 v = 2.42 ms-1

There is an external force being applied so A 10 kg mass is pulled along a rough horizontal surface by a horizontal force of magnitude 100N. The mass is initially at rest and it is pulled by the force for a distance of 10m. The coefficient of friction between the mass and the surface is 0.6. Use the work – energy principle to find the final speed of the mass. 0 ms-1 v ms-1 R Fr 100 N  = 0.6 10g 10 m There is an external force being applied so PEstart + KEstart + Win = PEend + KEend+ Wout 0 + 0 + Fs = 0 + ½mv² + Frs 100 x 10 = ½ x 10gv² + Frs 1000 = 5gv² + 10 x 6g 1000 – 60g = 5gv² v² = 412 v = 9.08 ms-1 (3 s.f.) We need to find friction Fr= R = 0.6 x 10g = 6g

PEstart + KEstart + Win = PEend + KEend+ Wout A female athlete of mass 65kg runs up a slope inclined a 5º to the horizontal. Her speed at the bottom of the slope is 4 ms-1 and her speed decreases to 3 ms-1 at the top of the slope. The slope is 120 m long and the resistances to motion can be assumed constant at 40 N. Determine the constant force she must exert to get to the top of the slope. 3 ms-1 R F 120 m 40 N 4 ms-1 65g R h = 120sin 5º F 40 N 5º zero PE level 65g The athlete starts with KE and work and ends with KE, PE and resistances PEstart + KEstart + Win = PEend + KEend+ Wout 0 + (½ x 65gx 4²) + 120F = (65g x 120sin 5º) + (½ x 65gx 3²) + (40 x 120) 520 + 120F = 6662.2 + 292.5 + 4800 120F = 11234.7 F = 93.6 N (3 s.f.)

Your turn now Blue book pg 261, 262 Mechanics 2 exercise 4B page 86 all the questions