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Work and Power 
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#1
Feb409, 07:53 PM

P: 5

1. The problem statement, all variables and given/known data
A car of mass 840.0 kg accelerates away from an intersection on a horizontal road. When the car speed is 37.4 km/hr (10.4 m/s), the net power which the engine supplies is 4300.0 W (in addition to the extra power required to make up for air resistance and friction). Calculate the acceleration of the car at that time. 2. Relevant equations Work=Force*Distance Power=Work/Time Force=Mass*Acceleration 3. The attempt at a solution I have tried numerous ways of solving this problem and remain unsuccessful. My closest attempt was using the equation Power=Kinetic Energy/Time, I used the velocity as Kinetic Energy to find the time and plugged that into the Kinematic equation Vf=Vi+a*d but that did not work either. Please help me! 


#2
Feb409, 09:49 PM

HW Helper
P: 3,394

Do you know some calculus?
It looks like a calculus problem to me. You can't use Power = KE/time: consider the situation where the car is moving at zero acceleration and a large speed. The power required ("extra" not counted as stated in problem) is zero but the KE is large. Instead, you should use Power = dE/dt, the derivative of kinetic energy with respect to time. The differentiation is not difficult, but does involve the chain rule. 


#3
Feb509, 09:04 AM

P: 5

yes I do know calculus, I am in Calc 2 but I still don't follow.



#4
Feb509, 09:34 AM

P: 56

Work and Power
M= 840.0 kg
P= 4300.0 W v= 10.4 m/s s = v*dt [you are working with power (which is watt per second). what should dt be?] gives s p = w/dt gives w w = f*s gives f f = m*a 


#5
Feb509, 11:45 AM

P: 5

how am I supposed to find dt when im not given a time



#6
Feb509, 11:57 AM

HW Helper
P: 5,343

As I recall power is equal to force times velocity.
Units of Watts = Nm/s P = F*V So the instantaneous power is given as 4300 w and the velocity is 10.4 m/s. That makes F = 4300/10.4 doesn't it? And you have the mass, so ... 


#7
Feb509, 11:59 AM

P: 5

i thought that power was equal to F*V but if you do the math it delivers units of Kg*m/s and not Kg*m/s^{2}. maybe im using the wrong velocity?



#8
Feb509, 12:33 PM

P: 116

Watt = J/s = kgm^{2}/s^{3} = Nm/s
So it works out just fine. :) 


#9
Feb509, 07:51 PM

HW Helper
P: 3,394

Oh, clever to remember P = Fv !
Just for interest, I was thinking of P = dE/dt = d/dt(.5mv^2) = .5m*2v*dv/dt = mva which is the same as P = Fv. 


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