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

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SUMMARY

The resistive force acting on a car coasting at a constant speed of 72 km/h on a decline can be calculated using the conservation of energy principle. Given a mass of 1600 kg and a height decrease of 200 m over 6000 m, the gravitational potential energy lost is 3.2 x 10^6 J. This energy is equal to the work done by the resistive force, leading to a calculated resistive force of 533 N. The speed of the car does not factor into this calculation since the kinetic energy remains constant.

PREREQUISITES
  • Understanding of gravitational potential energy (U = mgh)
  • Knowledge of work-energy principle (W = F s)
  • Basic trigonometry for calculating forces on inclines
  • Familiarity with Newton's laws of motion
NEXT STEPS
  • Study the relationship between gravitational force and incline angles using trigonometric functions
  • Learn about the conservation of energy in mechanical systems
  • Explore the concept of resistive forces in different contexts, such as friction and air resistance
  • Investigate the effects of varying mass and incline on resistive forces in physics problems
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Students studying physics, particularly those focusing on mechanics and energy conservation, as well as educators looking for practical examples of force calculations in real-world scenarios.

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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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