Question

The velocity of a body of mass $20kg$ decreases from $20m/s$ to $5m/s$ in a distance of $100m$. Force on the body is

• A. $-27.5N$
• B. $-47.5N$
• C. $-37.5N$
• D. $-67.5N$

Answer 1

The correct option is C

$-37.5N$

Step 1: Given data:

Mass of the body $m=20kg$

Initial velocity $u=20m/s$

Final velocity $v=5m/s$

Distance covered $s=100m$

Step 2: Formula used:

According to the third law of motion,

$v^2=u^2+2as\dots \dots \dots \dots \dots \dots \dots \dots \left(1\right)$

Where, $v$ is the final velocity, $u$ is the initial velocity, $a$ is the acceleration, $s$ is the distance covered

From Newton's second law,

$F=ma\dots \dots \dots \dots \dots \dots \dots \dots \dots \left(2\right)$

Where, $m,a$ are mass of the body and acceleration respectively.

Step 3: Calculation of acceleration:

From equation (1), the acceleration can be calculated as-

$v^2=u^2+2as$

Substituting the known values,

$\begin{array}{rcl}{5}^2& =& {20}^2+2a\times \left(100\right)\\ 25& =& 400+200a\\ -375& =& 200a\\ a& =& -1.875m/s^2\end{array}$

Thus, the retardation comes to be $-1.875m/s^2$.

Step 4: Calculation of force:

The force can be calculated using the formula in equation (2),

$$\begin{gathered} F=ma \\ F=\left(20\right)\times \left(-1.875\right) \\ F=-37.5N \end{gathered}$$

Thus, the force on the body is $-37.5N$.

Hence, option (C) is the correct answer.