Driving Force Calculation For A Car Ascending A Slope A Physics Problem

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A car with a mass of 1000 kg drives up a slope that is 750 meters long. The slope is inclined at an angle $\alpha$, where $\sin \alpha = \frac{1}{25}$. Determine the driving force exerted by the car's engine, given that the car's speed at the bottom of the incline is 25 m/s and its speed at the top is 15 m/s.

Keywords

  • Driving force
  • Inclined plane
  • Work-energy theorem
  • Gravitational potential energy
  • Kinetic energy

Solution

To determine the driving force of the engine, we need to analyze the forces acting on the car and apply the work-energy theorem. The work-energy theorem states that the net work done on an object is equal to the change in its kinetic energy.

1. Identify the Forces Acting on the Car

Several forces are acting on the car as it drives up the slope:

  • Driving Force ($F_d$): This is the force exerted by the car's engine, which propels the car up the slope. This is the force we want to calculate.
  • Gravitational Force ($F_g$): This force acts vertically downwards and is equal to the car's weight ($mg$), where $m$ is the mass (1000 kg) and $g$ is the acceleration due to gravity (approximately 9.8 m/s²). We can resolve this force into two components:
    • Component along the slope ($F_{g\parallel}$): This component acts downwards along the slope and is equal to $mg \sin \alpha$. This opposes the car's motion.
    • Component perpendicular to the slope ($F_{g\perp}$): This component acts perpendicular to the slope and is balanced by the normal reaction force from the road.
  • Normal Reaction Force ($N$): This force acts perpendicular to the slope and is exerted by the road on the car. It balances the component of the gravitational force perpendicular to the slope.
  • Frictional Forces ($F_f$): For simplicity, we will assume that frictional forces (air resistance and rolling friction) are negligible in this problem. If they were significant, they would act opposite to the direction of motion and would need to be included in the calculations.

2. Calculate the Component of Gravitational Force Along the Slope

Using the given values, we can calculate the component of gravitational force acting along the slope:

Fg∥=mgsin⁡α=(1000 kg)(9.8 m/s2)(125)=392 NF_{g\parallel} = mg \sin \alpha = (1000 \text{ kg})(9.8 \text{ m/s}^2)\left(\frac{1}{25}\right) = 392 \text{ N}

This force opposes the car's motion up the slope.

3. Apply the Work-Energy Theorem

The work-energy theorem states that the net work done on an object is equal to its change in kinetic energy. The net work done on the car is the work done by the driving force minus the work done by the component of gravity along the slope.

The work-energy theorem is a fundamental principle in physics that connects the concepts of work and energy. It is a powerful tool for analyzing motion, especially when dealing with situations where forces are not constant or the motion is not uniform. Understanding and applying the work-energy theorem is crucial for solving a wide range of physics problems. In our case, the work-energy theorem provides a direct link between the forces acting on the car, the distance it travels, and the change in its speed.

The work done by a force is given by the force multiplied by the distance over which it acts. Therefore:

  • Work done by the driving force ($W_d$) = $F_d \cdot d$, where $d$ is the distance traveled along the slope (750 m).
  • Work done by the gravitational force component ($W_g$) = $F_{g\parallel} \cdot d$, and since this force opposes the motion, the work is negative.

The change in kinetic energy ($\Delta KE$) is given by:

ΔKE=12m(vf2−vi2)\Delta KE = \frac{1}{2}m(v_f^2 - v_i^2)

where $v_f$ is the final velocity (15 m/s) and $v_i$ is the initial velocity (25 m/s).

Applying the work-energy theorem:

Wnet=ΔKEW_{net} = \Delta KE

Wd−Wg=ΔKEW_d - W_g = \Delta KE

Fd⋅d−Fg∥⋅d=12m(vf2−vi2)F_d \cdot d - F_{g\parallel} \cdot d = \frac{1}{2}m(v_f^2 - v_i^2)

4. Calculate the Change in Kinetic Energy

ΔKE=12(1000 kg)((15 m/s)2−(25 m/s)2)=12(1000)(−400) J=−200,000 J\Delta KE = \frac{1}{2}(1000 \text{ kg})((15 \text{ m/s})^2 - (25 \text{ m/s})^2) = \frac{1}{2}(1000)(-400) \text{ J} = -200,000 \text{ J}

The negative change in kinetic energy indicates that the car's kinetic energy decreased as it moved up the slope.

5. Substitute Values and Solve for the Driving Force

Now, we substitute the known values into the work-energy equation:

Fd(750 m)−(392 N)(750 m)=−200,000 JF_d(750 \text{ m}) - (392 \text{ N})(750 \text{ m}) = -200,000 \text{ J}

750Fd−294,000=−200,000750F_d - 294,000 = -200,000

750Fd=94,000750F_d = 94,000

Fd=94,000750≈125.33 NF_d = \frac{94,000}{750} \approx 125.33 \text{ N}

Therefore, the driving force of the engine is approximately 125.33 N.

6. Discuss the Result

The driving force calculated, approximately 125.33 N, is significantly less than the component of gravitational force acting along the slope (392 N). This means that while the engine is providing a force to propel the car upwards, a significant portion of the engine's effort is being used to counteract gravity. The negative change in kinetic energy indicates that the car is slowing down as it ascends the slope, which is expected given the opposing force of gravity and the relatively small driving force.

This problem highlights the interplay between different forms of energy – kinetic energy, gravitational potential energy, and the work done by external forces. The work-energy theorem provides a clear framework for understanding how these energies interact and how forces cause changes in the motion of an object. Understanding these concepts is crucial in understanding mechanical systems and problem-solving in physics.

It's also important to note that we made simplifying assumptions, such as neglecting friction. In a real-world scenario, friction would play a more significant role, increasing the driving force required to maintain a given speed or achieve a certain change in kinetic energy. Factors like air resistance, rolling resistance, and the efficiency of the engine all contribute to the overall forces acting on the car and the energy required for its motion.

Furthermore, the problem can be approached using other physics principles, such as Newton's Second Law of Motion. Applying Newton's Second Law along the slope would lead to the same result, providing an alternative perspective on the problem. This highlights the interconnectedness of various concepts in physics and the ability to approach problems from different angles. A comprehensive understanding of these principles enhances problem-solving skills and deepens the understanding of physical phenomena.

7. Gravitational Potential Energy Considerations

Another way to think about this problem is in terms of gravitational potential energy. As the car moves up the slope, its gravitational potential energy increases. The change in gravitational potential energy ($\Delta U$) is given by:

ΔU=mgh\Delta U = mgh

where $h$ is the change in height. Since $\sin \alpha = \frac{h}{d}$, we have $h = d \sin \alpha = 750 \text{ m} \cdot \frac{1}{25} = 30 \text{ m}$.

Thus,

ΔU=(1000 kg)(9.8 m/s2)(30 m)=294,000 J\Delta U = (1000 \text{ kg})(9.8 \text{ m/s}^2)(30 \text{ m}) = 294,000 \text{ J}

The total work done by the driving force must overcome the increase in gravitational potential energy and account for the decrease in kinetic energy. This provides another perspective on the energy transformations occurring in the system.

Conclusion

The driving force exerted by the car's engine is approximately 125.33 N. This force is required to overcome the component of gravitational force acting along the slope and to account for the decrease in the car's kinetic energy as it moves uphill. The problem illustrates the application of the work-energy theorem and the importance of considering the forces acting on an object and their effect on its motion and energy. The problem also sheds light on the significance of understanding energy transformations, such as the conversion of kinetic energy and the gain in gravitational potential energy, and the interplay between these different forms of energy in physical systems. This understanding is crucial for analyzing and solving a broad range of physics problems related to motion and forces.