# What does the Force in the equation P=FV represent?

So this is the question in context.

A car of mass 1200 kg tows a trailer of mass 300 kg along a straight horizontal road. The resistance to motion is modeled as being a constant force of magnitude 800 N acting on the car and a constant force of magnitude 200 N acting on the trailer. The power generated by the engine is 30 kW. Calculate the acceleration when the car is travellng at 12 ms$^{-1}$."

I understand the first part which is considering the car and trailer as one object, so a mass of 1500 Kg experiencing frictional force of 1000N. Now here's the part I don't get. I use the equation P=FV where P=30,000W and V=12m/s to get F=2500N. Does this 2500N represent the resultant force needed to get a velocity of 12 m/s (hence a forward force of 3500N, acceleration of 5/3 m/s/s) or does it represent purely the force of the engine required to get a velocity of 12 m/s (hence a resultant force of 1500N, acceleration of 1 m/s/s).

If the second option is true then does this mean resistance has no effect on the velocity?

• Only some of the power generated by the engine goes into accelerating the vehicle The rest goes into fighting friction. – David Hammen Jan 30 '18 at 23:20

## 1 Answer

The power required to sustain $v$ against friction is:

$$P_0 = F_0 v = 1000N \times 12 \frac{m}{s} = 12 kW$$

The excess power accelerates the vehicle:

$$P_x = P-P_0 = (Ma)v,$$

where $M=1500\,$kg is the total mass of the vehicle (and $P=30\,$kW).

• Do you mean 1000N? Also surely the second equation Px=P−P0=(Ma)v gives the force of the engine not the resultant force? – IK-_-IK Jan 30 '18 at 23:15
• @IK-_-IK Yes, fixed it in the post. Thx. I don't know what you mean by force of the engine versus resultant force. I'm saying some fraction of the power fights friction, and the complementary part accelerates at $a$ a mass $M$ which requires a force $F_a = Ma$, and at a speed $v$, that requires a power $F_av$, thereby determining $a$. – JEB Jan 31 '18 at 0:29