# Car Work and Power Calculation

1. Feb 4, 2009

### g.uricchio

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. Feb 4, 2009

### Delphi51

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. Feb 5, 2009

### g.uricchio

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

4. Feb 5, 2009

### timon

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. Feb 5, 2009

### g.uricchio

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

6. Feb 5, 2009

### LowlyPion

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 ...

7. Feb 5, 2009

### g.uricchio

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?

8. Feb 5, 2009

### Hannisch

Watt = J/s = kgm2/s3 = Nm/s

So it works out just fine. :)

9. Feb 5, 2009

### Delphi51

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.