MJackson09 Posted August 4, 2009 Posted August 4, 2009 Hey all, here is a problem that I have for homework and I can't figure it out. Any help will be appreciated. a 1650kg car accelerates from 15.9m/s to 28.1m/s in 8.31s. a.) what is the work done on the car? b.)what is the average power umed during acceleration? c.)what is the average force accelerating the car? So I know I have to use... Work = change in kinetic energy = 1/2 (m)v^2. Then what? Thanks ahead of time!
gonelli Posted August 4, 2009 Posted August 4, 2009 "change in kinetic energy" doesn't equal 1/2 (m)(v)^2 "kinetic energy" = 1/2 (m)(v)^2 Newton's equation (sum of)F = ma And the fact that "power" means "energy per unit of time" may also be useful
DJBruce Posted August 4, 2009 Posted August 4, 2009 (edited) Remember what change in kinetic energy actually means: [math]w=\Delta KE=(KE_{f}-KE_{i})=(\frac{1}{2}mv_{f}^{2})-(\frac{1}{2}mv_{i}^{2})[/math] Power is the rate of doing work so: [math]P=\frac{w}{t}[/math] To find force simply use Newtons second law: [math]F=ma[/math] You know m but you need to find a so simply use the definition of acceleration: [math]a=\frac{\Delta v}{\Delta t}= \frac{(v_{f}-v_{i})}{\Delta t} [/math] or if you want to do the first problem another way do it another way: [math]W=F\cdot d[/math] You know force and to find the distance simply remember your kinematic equations. Edited August 4, 2009 by DJBruce missed a parentheses
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