/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 51 A 1250 kg car is pulling a 325 k... [FREE SOLUTION] | 91Ó°ÊÓ

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A 1250 kg car is pulling a 325 kg trailer. Together, the car and trailer have an acceleration of \(2.15 \mathrm{m} / \mathrm{s}^{2}\) directly forward. a. Determine the net force on the car. b. Determine the net force on the trailer.

Short Answer

Expert verified
a. The net force on the car is \(2687.5 \mathrm{N}\). b. The net force on the trailer is \(698.75 \mathrm{N}\).

Step by step solution

01

Determine the net force on the car

The net force acting on the car can be determined by using Newton's second law, \( F = m \cdot a \). Here, \( m \) is the mass of the car and \( a \) is the acceleration. So by substituting the values into the formula, we get: \( F = 1250 \mathrm{kg} \cdot 2.15 \mathrm{m/s^{2}} = 2687.5 \mathrm{N}\).
02

Determine the net force on the trailer

Similarly, the net force acting on the trailer can also be determined by using Newton's second law, \( F = m \cdot a \). Here, \( m \) is the mass of the trailer and \( a \) is the acceleration. So by substituting the values into the formula, we get: \( F = 325 \mathrm{kg} \cdot 2.15 \mathrm{m/s^{2}} = 698.75 \mathrm{N}\).

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Key Concepts

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

Net Force Calculation
Understanding the net force calculation is crucial in physics as it directly influences an object's motion. In the given problem, we look at a practical application: a car towing a trailer and accelerating.

Net force is the vector sum of all forces acting on an object. If these forces are unbalanced, the object will accelerate, as explained by Newton's second law of motion: \( F = m \cdot a \) where \( F \) is the net force, \( m \) is the mass of the object, and \( a \) is its acceleration.

Applying Net Force Calculation

For the car and trailer scenario, we calculate the net force on each separately despite their connection, because they can experience different forces individually. By plugging the mass of the car and its acceleration into the formula, we obtain the net force on the car. Similarly, we apply the same process for the trailer. This calculation assumes all forces apart from the car’s engine force are negligible, focusing solely on the ‘forward’ direction indicated by the acceleration.
Physics Problem Solving
Solving physics problems elegantly hinges on a clear understanding of the underlying concepts and a methodical approach to apply them to real-world scenarios.

When addressing problems like the one involving a car and trailer, it's important to:
  • Identify the known variables (e.g., mass and acceleration)
  • Determine the appropriate physical law (Newton's second law)
  • Apply the law systematically to find the unknown quantity (net force)

Implementing a Systematic Approach

In our example, we dissected the problem into two parts, calculating the net force on the car first, then on the trailer. This step-by-step approach turns a complex scenario into manageable chunks. Also, thinking about each object independently simplifies the complexity of the system. By making sure to conceptualize the problem correctly, one reduces errors and ensures accuracy in the calculations.
Classical Mechanics
Classical mechanics forms the foundation for understanding the motion of objects, from daily life scenarios to planetary motion. It encompasses several laws and principles, with Newton's laws of motion at its core.

In the case of the car and the trailer, we are witnessing classical mechanics in action. These laws enable us to predict

Motion and Forces

As the car accelerates, it exerts a force on the trailer, and the same force acts upon the car but in the opposite direction due to Newton's third law, often phrased as 'for every action, there is an equal and opposite reaction.' Although this reactive force isn't part of the net force calculation for the car, it is a necessary consideration in understanding the interaction between the two objects.

The beauty of classical mechanics lies in its ability to quantify these natural phenomena, presenting them in a formulaic and predictable pattern, making it an essential part of physics that enables us to comprehend and describe the workings of the universe on a macroscale.

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Most popular questions from this chapter

A 0.150 kg baseball is thrown upward with an initial speed of \(20.0 \mathrm{m} / \mathrm{s}\) a. What is the force on the ball when it reaches half of its maximum height? (Disregard air resistance. \()\) b. What is the force on the ball when it reaches its peak?

See Sample Problem \(B\) Two lifeguards pull on ropes attached to a raft. If they pull in the same direction, the raft experiences a net force of \(334 \mathrm{N}\) to the right. If they pull in opposite directions, the raft experiences a net force of \(106 \mathrm{N}\) to the left. a. Draw a free-body diagram representing the raft for each situation. b. Find the force exerted by each lifeguard on the raft for each situation. (Disregard any other forces acting on the raft.)

A boat moves through the water with two forces acting on it. One is a \(2.10 \times 10^{3} \mathrm{N}\) forward push by the motor, and the other is a \(1.80 \times 10^{3} \mathrm{N}\) resistive force due to the water. a. What is the acceleration of the 1200 kg boat? b. If it starts from rest, how far will it move in 12 s? c. What will its speed be at the end of this time interval?

The force that attracts Earth to an object is equal to and opposite the force that Earth exerts on the object. Explain why Earth's acceleration is not equal to and opposite the object's acceleration.

Draw free-body diagrams showing the weight and normal forces on a laundry basket in each of the following situations: a. at rest on a horizontal surface b. at rest on a ramp inclined \(12^{\circ}\) above the horizontal c. at rest on a ramp inclined \(25^{\circ}\) above the horizontal d. at rest on a ramp inclined \(45^{\circ}\) above the horizontal

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