This **article aims** to find **acceleration imparted on the box** placed on a **frictionless table** on earth.

In **mechanics**, **acceleration is rate of change of an object’s velocity with respect to time.** Accelerations are vector quantities having both magnitude and direction. The **direction** of an object’s acceleration is given by orientation of the **net force acting** on that object. The **magnitude** of the object’s acceleration, as described by **Newton’s second law,** is the combined effect of two causes:

- The
**net balance of all external forces**acting on that object — the magnitude is**directly proportional**to this resulting resultant force - The
**weight of that object**, depending on the materials it is made of — size is inversely proportional to the**object’s mass.**

The **SI** unit is **meters per second squared**, $\dfrac{m}{s^{2}}$.

**Average Acceleration**

**Average acceleration** is the** rate of change of velocity** $\Delta v$ divided over the time $\Delta t$.

\[a=\dfrac{\Delta v}{\Delta t}\]

**Instantaneous Acceleration**

**Instantaneous acceleration** is the **limit of average acceleration** over an infinitesimally **small time interval.** Numerically, the instantaneous acceleration is the **derivative of the velocity vector with respect to time.**

\[a=\dfrac{dv}{dt}\]

Since **acceleration** is defined as the **derivative of velocity** $v$ with respect to time $t$ and velocity are defined as **derivative of position** $x$ with respect to time, **acceleration** can be thought of as **second derivative** of $x$ with respect to $t$:

\[a=\dfrac{dv}{dt}=\dfrac{d^{2}x}{d^{2}t}\]

** **

**Newton’s Second Law of Motion**

The proper acceleration, i.e., the **acceleration of the body relative to the state of free fall**, is measured by an **accelerometer.** In classical mechanics, for a body having constant mass (vector), the **acceleration of the body’s center of gravity** is **proportional to the net force vector** (i.e., the sum of all forces) acting on it **(Newton’s second law):**

\[F=ma\]

\[a=\dfrac{F}{m}\]

$F$ is the **net force acting on body**, and $m$ is the **mass.**

**Expert Answer**

**Data given in the question** is:

\[a(acceleration) of \: the \:block=5.3\dfrac{m}{s^{2}}\]

\[F(horizontal force)=10\:N\]

\[a(acceleration)\: due \:to\:gravity=1.62\dfrac{m}{s^{2}}\]

The **value of mass** is calculated by using the following formula:

\[F=\dfrac{m}{a}\]

\[m=\dfrac{F}{a}\]

\[m=\dfrac{10}{5.3}\]

\[m=1.89\:kg\]

The mass of the box is $1.89\:kg$.

The **value of the acceleration** is found by using the following formula:

\[F=ma\]

\[a=\dfrac{F}{m}\]

\[a=\dfrac{5}{1.89}\]

\[a=2.65\dfrac{m}{s^{2}}\]

Hence, **acceleration imparted to the block** is $2.65\dfrac{m}{s^{2}}$.

**Numerical Result**

**Acceleration imparted to the block** is $2.65\dfrac{m}{s^{2}}$.

**Example**

The block is on a frictionless table on the ground. The block accelerates at $5\dfrac{m}{s^{2}}$ when acted upon by a horizontal force of $20\: N$. The block and table are placed on the moon. Gravitational acceleration on the surface of the Moon is $1.8\dfrac{m}{s^{2}}$.When the block is on the moon, a horizontal force of $15\:N$ acts on it.

**Solution**

**Data given in the example** is:

\[a(acceleration) of \: the \:block=5\dfrac{m}{s^{2}}\]

\[F(horizontal force)=20\:N\]

\[a(acceleration)\: due \:to\:gravity=1.8\dfrac{m}{s^{2}}\]

The **value of mass** is calculated by using the following formula:

\[F=\dfrac{m}{a}\]

\[m=\dfrac{F}{a}\]

\[m=\dfrac{20}{5}\]

\[m=4\:kg\]

The mass of the box is $4\:kg$.

The** value of the acceleration** is found by using the following formula:

\[F=ma\]

\[a=\dfrac{F}{m}\]

\[a=\dfrac{15}{4}\]

\[a=3.75\dfrac{m}{s^{2}}\]

Hence, **acceleration imparted to the block** is $3.75\dfrac{m}{s^{2}}$.