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Chapter 5: Problem 46

A car weighing 12,500 N starts from rest and accelerates to \(83.0 \mathrm{km} / \mathrm{h}\) in \(5.00 \mathrm{s}\). The friction force is 1350 N. Find the applied force produced by the engine.

Short Answer

Expert verified
The applied force produced by the engine can be found by first calculating the car's acceleration using the formula \( a = \frac{v_f - v_i}{t} \), converting the final velocity to m/s, and finding the mass of the car using \( m = \frac{W}{g} \). Then, using Newton's second law, \( F_\text{net} = m \cdot a \), and considering the friction force, we can solve for the applied force produced by the engine: \( F_\text{net} = F_\text{engine} - F_\text{friction} \).

Step by step solution

01

Calculate the car's acceleration

To find the acceleration, we will first convert the final velocity from km/h to m/s, and then use the following equation, knowing the initial velocity is 0 as the car starts from rest: \( a = \dfrac{v_f - v_i}{t} \)
02

Convert the Final Velocity to m/s

To convert the final velocity (83.0 km/h) to m/s, we can use the following conversion factor: 1 km/h = 1000 m / 3600 s So, \(v_f\) in m/s will be: \(v_f = 83.0\ \text{km/h} \cdot \dfrac{1000\ \text{m}}{3600\ \text{s}}\)
03

Calculate the Acceleration

Now, we can substitute the converted final velocity, initial velocity (0) and the given time into the acceleration equation: \( a = \dfrac{ (v_f \cdot \dfrac{1000}{3600}) - 0}{5.00\ \text{s}}\)
04

Apply Newton's Second Law

According to Newton's second law, the net force (F_net) acting on an object is given by the product of its mass (m) and acceleration (a): \( F_\text{net} = m \cdot a \) Since we know the weight of the car (12,500 N) and the gravitational acceleration (g = 9.81 m/s虏), we can find the mass of the car with the formula: \( m = \dfrac{W}{g} \) Then we can find the net force acting on the car by plugging in the mass and calculated acceleration.
05

Calculate the Applied Force by the Engine

We know the friction force (1350 N) acting on the car. The net force is the difference between the applied force produced by the engine (F_engine) and the friction force, as they act in opposite directions: \( F_\text{net} = F_\text{engine} - F_\text{friction} \) Now, we can solve for the applied force produced by the engine (F_engine). This step-by-step breakdown of the problem provides a clear guide on how to calculate the applied force produced by the engine of the car.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Acceleration Calculation
To determine acceleration, it's crucial first to convert the velocity from kilometers per hour (km/h) to meters per second (m/s). This is because the standard SI unit for velocity is meters per second. For instance, to convert 83 km/h to m/s, we use the conversion factor: 1 km/h is equal to approximately 0.27778 m/s. Thus, we have:
  • Final velocity, \( v_f = 83.0 \text{ km/h} \times 0.27778 \text{ m/s per km/h} \)
  • This gives us \( v_f \approx 23.06 \text{ m/s} \)
Next, we use the acceleration formula, \( a = \frac{v_f - v_i}{t} \). Since the car starts from rest, the initial velocity \( v_i = 0 \). The time given is 5 seconds. Hence, the acceleration \( a \) becomes:
  • \( a = \frac{23.06 \text{ m/s} - 0}{5 \text{ s}} \)
  • \( a \approx 4.612 \text{ m/s}^2 \)
Friction Force
Friction is the force resisting the relative motion of solid surfaces sliding against each other. In this problem, the friction force acting on the car is given as 1350 N. This is an opposing force that the engine needs to overcome in addition to accelerating the car.
Understanding friction is important because it plays a significant role in real-world motion problems. Here:
  • The friction force \( F_{\text{friction}} \) = 1350 N
This force is essentially a hindrance to the engine's efforts to move the car forward. It impacts the net force that actually accelerates the car, which will be considered when calculating the necessary engine force.
Kinematic Equations
Kinematic equations describe the motion of objects using variables such as time, displacement, velocity, and acceleration. These equations form the backbone for solving many physics problems.
  • In this exercise, the primary kinematic equation used is \( a = \frac{v_f - v_i}{t} \)
This equation provides a straightforward method to compute acceleration given initial and final velocities and the time of motion. Since the car starts from a standstill, the initial velocity is zero, simplifying the calculations.
The kinematic equations are essential for understanding how different physical quantities relate and influence each other in the context of motion.
Force and Motion
The relationship between force and motion is fundamental in physics and is described by Newton's Second Law of Motion. This law states that the net force acting on an object is equal to the product of its mass and acceleration:
  • \( F_{\text{net}} = m \cdot a \)
For this problem:
  • The weight of the car, 12,500 N, gives us its mass, \( m = \frac{W}{g} \), where \( g = 9.81 \text{ m/s}^2 \)
  • \( m = \frac{12500}{9.81} \approx 1274.2 \text{ kg} \)
  • Using the calculated acceleration, \( a \approx 4.612 \text{ m/s}^2 \), the net force \( F_{\text{net}} \) is \( 1274.2 \times 4.612 \approx 5879.1 \text{ N} \)
Applied Force Calculation
To find the applied force by the engine, we need to account for both the net force and the effect of friction. Using the relationship:
  • \( F_{\text{net}} = F_{\text{engine}} - F_{\text{friction}} \)
We rearrange this to solve for the engine force:
  • \( F_{\text{engine}} = F_{\text{net}} + F_{\text{friction}} \)
  • Substitute the known values: \( F_{\text{engine}} = 5879.1 \text{ N} + 1350 \text{ N} \)
  • Thus, \( F_{\text{engine}} \approx 7229.1 \text{ N} \)
This calculation shows the actual force the engine needs to exert to not only overcome friction but also to accelerate the car as described.

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Most popular questions from this chapter

An 80.0-kg passenger in an SUV traveling at \(1.00 \times 10^{3} \mathrm{km} / \mathrm{h}\) is wearing a seat belt. The driver slams on the brakes and the SUV stops in \(45.0 \mathrm{m}\). Find the force of the seat belt on the passenger.

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