Question

The engine of a car produces an acceleration of $6\dfrac{m}{{{s^2}}}$ in the car. If this car pulls a block of the same mass, then the acceleration would be(in $\dfrac{m}{{{s^2}}}$):
a. $6$
b. $12$
c. $3$
d. $1.5$

Answer 1

Hint: Before we proceed into the problem, it is important to know about the formula of force. Find out the expression for force relating mass of an object. The force is simply the mass of an object multiplied by the acceleration of the body.

Complete answer:
Given that the engine of a car produces an acceleration of $6\dfrac{m}{{{s^2}}}$ in the car and the car pulls another car of same mass.

Force is an external agent capable of changing the state of rest or motion of a body. It has a magnitude and a direction. The direction towards which the force is applied is known as the direction of the force and the application of force is the point where force is applied. The quantity of force is expressed by the vector product of mass (m) and acceleration (a).

The equation or the formula for force can mathematically be expressed in the form of:
$F = m\vec a$
Where a = acceleration of body
m = mass of the body

Substitute the values,
$ \Rightarrow F = m \times 6$
$ \Rightarrow \dfrac{F}{6} = m$------ (1)

As the mass of the car is 'm' and the car is pulling another car of same mass i.e., 'm'. Therefore, the new mass is 2m (As engine has to pull both).
$ \Rightarrow F = 2m \times a$

Substituting m value from equation 1
$ \Rightarrow F = 2 \times \dfrac{F}{6} \times a$
$ \Rightarrow a = 3\dfrac{m}{{{s^2}}}$
Therefore, the acceleration of the car is 3 $\dfrac {m}{s^2}$.

So, option (c) is correct.

Note: The magnitude of force will be same, it’s just the mass is doubled. If the mass is doubled keeping force constant, the acceleration is halved. Keep in mind not to double the magnitude of force it can alter the answer we are expecting.