Question #6352

A car with a mass of 1.50x10^3 kg starts from rest and accelerates to a speed of 18.0 m/s in 12.0s. Assume that the force of resistance constant at 400.0 N during this time. What is the average power developed by the car's engine?

Expert's answer

From the equation for speed V = a*t let's find car's acceleration:

a = V/t.

Using the Newton's Second Law:

F - Fres = ma,

where Fres is the force of

resistance (Fres = 400 N),

F - force of engine. So, F = Fres + ma.

Average

power can be found from the equations for total engine's work:

A = F*S = P*t,

P = F*S/t.

S can be found from the equation of displacement:

S =

(a*t^2)/2.

So, P = (Fres + m*V/t)*(V/t*t^2)/(2t) = (Fres + m*V/t)*V/2 =

(400+1500*18/12)*18/2 = 23850 W.

a = V/t.

Using the Newton's Second Law:

F - Fres = ma,

where Fres is the force of

resistance (Fres = 400 N),

F - force of engine. So, F = Fres + ma.

Average

power can be found from the equations for total engine's work:

A = F*S = P*t,

P = F*S/t.

S can be found from the equation of displacement:

S =

(a*t^2)/2.

So, P = (Fres + m*V/t)*(V/t*t^2)/(2t) = (Fres + m*V/t)*V/2 =

(400+1500*18/12)*18/2 = 23850 W.

## Comments

Assignment Expert20.06.13, 15:03It is negative, the sign here means direction. We have F-Fres=ma, so F and Fres have opposite directions, F is collinear with direction of the car motion, Fres is opposite to it. The magnitude of Fres is positive, that's right, but when construction equations we keep in mind that it's directed opposite to the motion of the car.

Philo20.06.13, 01:32But the friction force is supposed to be negative

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