What Is the Resistive Force Acting on a Car Coasting at Constant Speed?

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The discussion centers on calculating the resistive force acting on a car coasting at a constant speed of 72 km/h on a decline. The car, with a mass of 1600 kg, loses gravitational potential energy as it descends 200 m over 6000 m, resulting in a potential energy loss of 3.2 million joules. This energy loss is equated to the work done by the resistive force, leading to a calculated force of 533 N. The accuracy of using gravitational acceleration as 10 m/s² instead of 9.8 m/s² is confirmed, and the kinetic energy is deemed irrelevant since the car maintains constant speed. The resistive force is ultimately determined to be equal to the gravitational force acting on the car on the incline.
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Homework Statement


Along a certain section of road a car (m=1600) will coast in neutral at a constant speed of 72km/h if there is no wind.Examination of a topological map shows that for this section of the road the elevation decreases by 200m for each 6000m of road.What is the resistive force that acts on the car when it is traveling at 72km/h?

2. The attempt at a solution
I applied the conservation of energy and i said. For every 6000m the car is falling by 200m so is losing gravitational potential energy equal with U=mgh=1600 \cdot 10 \cdot 200 =3.2 10^6
We know the car has constant speed so this energy didnt become kinetic thus has to be equal with the work of the resistive force which is W=F s and s=6000m

So we get the equation U=W \rightarrow mgh=F s \rightarrow F=\frac{mgh}{s}

So F=\frac{3.2 \cdot 10^6}{6 \cdot 10^3} \rightarrow F=533 NCould you please someone tell me if i did something wrong and where? I didnt use at all the speed of the car since i didnt use the kinetic energy. Thank you
 
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mlazos said:
Could you please someone tell me if i did something wrong and where? I didnt use at all the speed of the car since i didnt use the kinetic energy.
What you did is fine (assuming you're using g = 10 m/s^2--9.8 m/s^2 might be more accurate). The KE doesn't change so it's irrelevant.

Another way to view things is to just view the forces acting on the car directly. Since it's on a decline, there's a gravitational force acting on it equal to mg \sin\theta = mgh/s. Since the car's not accelerating, the resistive force must equal that gravitational force.
 
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