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The engine of a car produces an acceleration of \(6 \mathrm{~ms}^{-2}\) in the car. If this car pulls another car of the same mass, then the acceleration would be (a) \(6 \mathrm{~ms}^{-2}\) (b) \(12 \mathrm{~ms}^{-2}\) (c) \(3 \mathrm{~ms}^{-2}\) (d) \(1.5 \mathrm{~ms}^{-2}\)

Short Answer

Expert verified
The acceleration would be \(3 \mathrm{~ms}^{-2}\).

Step by step solution

01

Understand the Problem

The problem states that a car engine produces an acceleration of \(6 \mathrm{~ms}^{-2}\) when moving by itself. When it pulls another car of the same mass, we need to determine the new acceleration of the system.
02

Consider the Effect of Doubling the Mass

When the car pulls another car of the same mass, the total mass the engine has to move doubles. According to Newton's Second Law, \( F = ma \), if the force \( F \) remains constant, doubling the mass \( m \) will result in halving the acceleration \( a \).
03

Calculate the New Acceleration

Since the mass has doubled, the acceleration will be half of the original. Thus, the new acceleration will be \( \frac{6}{2} = 3 \mathrm{~ms}^{-2} \).

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Key Concepts

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

acceleration
Acceleration is a fundamental concept in physics, describing how quickly an object's velocity changes over time. It is a vector quantity, meaning it has both magnitude and direction. Acceleration occurs whenever an object speeds up, slows down, or changes direction. The standard unit of acceleration in physics is meters per second squared (\( \mathrm{ms}^{-2} \)).
When we talk about an engine producing an acceleration, like in the car problem, it involves the engine increasing the car's speed over time. In the given scenario, the engine acceleration is initially \( 6 \mathrm{~ms}^{-2} \), which means every second, the car's speed increases by 6 meters per second. Understanding acceleration helps us predict how the speed of an object changes, which is essential for calculating distances and movements in various contexts.
force and mass relationship
One of the key principles of physics is the relationship between force, mass, and acceleration, as outlined by Newton's Second Law of Motion. According to this law, the net force acting on an object is equal to the mass of the object times its acceleration:\[ F = ma \]This means if you know any two of the values — force (\( F \)), mass (\( m \)), or acceleration (\( a \)) — you can calculate the third.
In the example of the car, as the car pulls another vehicle of the same mass, the total mass the engine needs to move is doubled. With the mass doubling, if the force produced by the engine remains constant, the acceleration must decrease to maintain this relationship. When force is constant and mass increases, acceleration must decrease inversely. That's why pulling an additional car reduces the acceleration from \( 6 \mathrm{~ms}^{-2} \) to \( 3 \mathrm{~ms}^{-2} \).
This relationship is crucial in understanding how changes in mass and force affect acceleration in practical scenarios, such as calculating the effects of added weight on vehicles or understanding the impacts of forces on various objects.
problem-solving in physics
To effectively solve physics problems like the car's acceleration question, there are some key strategies and steps to follow:
  • Read the problem carefully: Understanding what is being asked is crucial. Identify the given values, ask what needs to be found, and note any important words like 'double' or 'constant'.
  • Identify relevant principles: Determine which physics laws or concepts apply to the situation. In this case, it’s Newton’s Second Law, which connects force, mass, and acceleration.
  • Use equations: Write the applicable formula(s) and insert the known values. Here, the formula \( F = ma \) will guide you to understand how acceleration changes with mass.
  • Calculate and interpret: Solve the equation step by step and interpret your findings in the context of the question. Double-check your results to ensure they're logical.
Solving physics problems becomes much easier with practice, ensuring you understand the underlying concepts and can apply them systematically. Structured problem-solving helps develop a deeper comprehension of physics, preparing you for more complex scenarios and questions you might encounter.

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Most popular questions from this chapter

A player caught a cricket ball of mass \(150 \mathrm{~g}\) moving at a rate of \(20 \mathrm{~ms}^{-1}\), If the catching process is completed in \(0.1 \mathrm{~s}\), the force of blow exerted by the ball on the hands of the player is equal to [AIFEE 2006] (a) \(3 \bar{N}\) (b) \(-30 \mathrm{~N}\) (c) \(300 \mathrm{~N}\) (d) \(150 \mathrm{~N}\)

A block of mass \(2 \mathrm{~kg}\) is placed on the floor. The coefficient of static friction is \(0.4\). If a force of \(2.8 \mathrm{~N}\) is applied on the block parallel to floor, the force of friction between the block and floor (Taking \(\left.g=10 \mathrm{~ms}^{-2}\right)\) is (a) \(2.8 \mathrm{~N}\) (b) \(8 \mathrm{~N}\) (c) \(2 \mathrm{~N}\) (d) zero

A cricket ball of mass 150 g collides straight with a bat with a velocity of \(10 \mathrm{~ms}^{-1}\). Batsman hits it straight back with a velocity of \(20 \mathrm{~ms}^{-1}\), If ball remains in contact with bat for \(0.18\), then average force exerted by the bat on the ball is (a) \(15 \mathrm{~N}\) (b) \(45 \mathrm{~N}\) (c) \(150 \mathrm{~N}\) (d) \(4.5 \mathrm{~N}\)

A bomb of mass \(16 \mathrm{~kg}\) at rest explodes into two pieces of masses \(4 \mathrm{~kg}\) and \(12 \mathrm{~kg}\). The velocity of the \(12 \mathrm{~kg}\) mass is \(4 \mathrm{~ms}^{-1}\). The kinetic energy of the other mass 18 (a) \(288 \mathrm{~J}\) (b) \(192 \mathrm{~J}\) (c) \(96 \mathrm{~J}\) (d) \(144 \mathrm{~J}\)

A \(40 \mathrm{~kg}\) slab rests on a frictionless floor. A \(10 \mathrm{~kg}\) block rests on top of the slab. The static coefficient of friction between the block and the slab is \(0.60\) while the kinetic coefficient of friction is \(0.40\). The \(10 \mathrm{~kg}\) block is acted upon by a horizontal force of \(100 \mathrm{~N}\). If \(g=9.8 \mathrm{~ms}^{-2}\), the resulting acceleration of the slab will be (a) \(1.47 \mathrm{~ms}^{-2}\) (b) \(1.69 \mathrm{~ms}^{-2}\) (c) \(9.8 \mathrm{~ms}^{-2}\) (d) \(0.98 \mathrm{~ms}^{-2}\)

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