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4.00 X10^5 J of work are done on a 1208 kg car while it accelerates from 10.0 m/s to...
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By definition the work done on a body of mass `m` is equal to the variation of its total energy between final and initial states.
`W =Delta(E) =E_f-E_i`
In this case we assume that only the kinetic energy is varying (the height of the body does not change thus the variation of potential energy is zero).
The initial kinetic energy is
`Ek_i = (m*v_i^2)/2 =1208*10^2/2 =60400 J`
The final kinetic energy will be
`Ek_f = W+Ek_i =400000* +60400*10^5 =460400 J `
Since the final kinetic energy is
`Ek_f = (m*v_f^2)/2`
the final speed will be
`v_f = sqrt((2*Ek_f)/m) =sqrt(2*460400/1208) =27.609 m/s`
The final speed of the car is 27.609 m/s
Posted by valentin68 on October 26, 2013 at 5:55 PM (Answer #1)
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