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Ch 07: Newton's Third Law
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook
All textbooksKnight Calc 5th EditionCh 07: Newton's Third LawProblem 8a
Chapter 7, Problem 8a

A 1000 kg car is pushing an out-of-gear 2000 kg truck that has a dead battery. When the driver steps on the accelerator, the drive wheels of the car push horizontally against the ground with a force of 4500 N. Rolling friction can be neglected. What is the magnitude of the force of the car on the truck?

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Step 1: Start by identifying the total system mass. The car has a mass of 1000 kg, and the truck has a mass of 2000 kg. Therefore, the total mass of the system is the sum of the two: \( m_{\text{total}} = 1000 \; \text{kg} + 2000 \; \text{kg} = 3000 \; \text{kg} \).
Step 2: Use Newton's Second Law of Motion, \( F = ma \), to calculate the acceleration of the system. The net force acting on the system is the force exerted by the car's drive wheels, \( F = 4500 \; \text{N} \). The acceleration of the system is given by \( a = \frac{F}{m_{\text{total}}} \). Substitute the values: \( a = \frac{4500}{3000} \; \text{m/s}^2 \).
Step 3: Focus on the interaction between the car and the truck. The force of the car on the truck is the same as the force required to accelerate the truck alone. Use Newton's Second Law again, \( F_{\text{car on truck}} = m_{\text{truck}} \cdot a \), where \( m_{\text{truck}} = 2000 \; \text{kg} \) and \( a \) is the acceleration calculated in Step 2.
Step 4: Substitute the values into the equation from Step 3: \( F_{\text{car on truck}} = 2000 \cdot a \). Use the acceleration value from Step 2 to complete the calculation.
Step 5: Conclude that the magnitude of the force of the car on the truck is equal to the force required to accelerate the truck at the same rate as the system. This force is determined by the truck's mass and the system's acceleration.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Newton's Second Law of Motion

Newton's Second Law states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. This law can be expressed with the formula F = ma, where F is the net force, m is the mass, and a is the acceleration. In this scenario, understanding how the forces interact between the car and the truck is crucial for determining the force exerted by the car on the truck.
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Force Interaction

According to Newton's Third Law of Motion, for every action, there is an equal and opposite reaction. This means that when the car exerts a force on the truck, the truck exerts an equal force back on the car. This concept is essential for analyzing the forces at play in the system, particularly in understanding how the car's force affects the truck's motion.
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System of Objects

In this problem, the car and truck can be considered as a single system. The total mass of the system affects how the forces are distributed and how the system accelerates. By treating the car and truck together, we can apply Newton's laws to find the force exerted by the car on the truck, taking into account the combined mass and the force applied by the car.
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Related Practice
Textbook Question

Block A in FIGURE EX7.4 is sliding down the incline. The rope is massless, and the massless pulley turns on frictionless bearings, but the surface is not frictionless. The rope and the pulley are among the interacting objects, but you'll have to decide if they're part of the system. Draw an interaction diagram.

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Textbook Question

How much force does the astronaut exert on his chair while accelerating straight up at 10 m/s2?

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Textbook Question

The foot of a 55 kg sprinter is on the ground for 0.25 s while her body accelerates from rest to 2.0 m/s. What is the magnitude of the friction force?

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Textbook Question

The sled dog in FIGURE EX7.15 drags sleds A and B across the snow. The coefficient of friction between the sleds and the snow is 0.10. If the tension in rope 1 is 150 N, what is the tension in rope 2?

Textbook Question

Blocks with masses of 1 kg, 2 kg, and 3 kg are lined up in a row on a frictionless table. All three are pushed forward by a 12 N force applied to the 1 kg block. How much force does the 2 kg block exert on the 1 kg block?

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Textbook Question

Block A in FIGURE EX7.4 is sliding down the incline. The rope is massless, and the massless pulley turns on frictionless bearings, but the surface is not frictionless. The rope and the pulley are among the interacting objects, but you'll have to decide if they're part of the system. Draw a free-body diagram for each object in the system. Use dashed lines to connect members of an action/reaction pair.

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