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**Hint:**In the first case, the value of the mass of an object and value of the acceleration of an object is given and applied force is asked. Use Newton’s second law of motion i.e., ${\text{F = ma}}$ and then substitute the given values in the formula and evaluate the value of the applied force. Now in the second case, the value of applied force in both the cases are the same. Value of mass of an object is given. Again using Newton’s second law of motion evaluate the value of acceleration of an object.

**Complete step by step solution:**

Given: Mass of an object, ${{\text{m}}_{\text{1}}}{\text{ = 16 kg}}$

Acceleration of the object, ${{\text{a}}_{\text{1}}}{\text{ = 3 m/}}{{\text{s}}^{\text{2}}}$

To find: Applied force

Using Newton’s second law of motion

${\text{F = ma}}$

Where F = Applied force

M = Mass of the object

And a = Acceleration of the object

Now substituting the given values in above formula, we get

${{\text{F}}_{\text{1}}}{\text{ = }}{{\text{m}}_{\text{1}}}{{\text{a}}_{\text{1}}}{\text{ = 16 \times 3 = 48 N}}$

Given that applied force on first object is same as that of the applied force on the another object

$\therefore {\text{ }}{{\text{F}}_{\text{1}}}{\text{ = }}{{\text{F}}_{\text{2}}}{\text{ = 48 N}}$

Mass of another object, ${{\text{m}}_{\text{2}}}{\text{ = 24 kg}}$

Acceleration of the object =?

Again using Newton’s second law of motion

${\text{F = ma}}$

Where F = Applied force

M = Mass of the object

And a = Acceleration of the object

Now substituting the given values in above formula, we get

$

{F_2}{\text{ = }}{{\text{m}}_{\text{2}}}{{\text{a}}_{\text{2}}} \\

\Rightarrow {{\text{a}}_{\text{2}}}{\text{ = }}\dfrac{{{{\text{F}}_{\text{2}}}}}{{{{\text{m}}_{\text{2}}}}}{\text{ }} = {\text{ }}\dfrac{{48}}{{24}}{\text{ = 3 m/}}{{\text{s}}^2} \\

$

**Therefore, applied force on an object is 48 N and acceleration of another object is 3 $m/{s^2}$.**

**Note:**The second law states that the rate of change of momentum of a body is directly proportional to the force applied, and this change in momentum takes place in the direction of the applied force.

\[{\text{F = }}\dfrac{{{\text{dp}}}}{{{\text{dt}}}}{\text{ = }}\dfrac{{{\text{d(mv)}}}}{{{\text{dt}}}}{\text{ = m}}\dfrac{{{\text{dv}}}}{{{\text{dt}}}}{\text{ = ma}}\]

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