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Work and Power

by g.uricchio
Tags: power, work
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Feb4-09, 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




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!
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Feb4-09, 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.
Feb5-09, 09:04 AM
P: 5
yes I do know calculus, I am in Calc 2 but I still don't follow.

Feb5-09, 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
Feb5-09, 11:45 AM
P: 5
how am I supposed to find dt when im not given a time
Feb5-09, 11:57 AM
HW Helper
P: 5,343
As I recall power is equal to force times velocity.

Units of Watts = N-m/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 ...
Feb5-09, 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/s2. maybe im using the wrong velocity?
Feb5-09, 12:33 PM
P: 116
Watt = J/s = kgm2/s3 = Nm/s

So it works out just fine. :)
Feb5-09, 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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