Question

A block of mass $5kg$ is moving horizontally at a speed of $1.5\dfrac{m}{s}$. A perpendicular force of $5N$ acts on it for $4\sec $. What will be the displacement of the block from the point where the force started acting?

Answer 1

Hint: To solve this question, we need to understand the concept of displacement first. Displacement is defined as the change in position of an object. It is a vector quantity and has both direction and magnitude. The SI unit of displacement is metre.

Complete step by step answer:
Let us assume initial velocity of $1.5\dfrac{m}{s}$ is in the x-direction
Since in the horizontal direction, no forces are acting. So, we can say that there is no acceleration taking place.
We know that the formula for speed is,
$v = \dfrac{d}{t}$
$d = v \times t$
So the displacement in the x direction is,
${S_X} = 1.5 \times 4$
${S_X} = 6m$
In the y direction,
$F = 5N$
$m = 5kg$
So, the acceleration in the y direction is,
${a_y} = \dfrac{F}{m}$ on putting the values,
${a_y} = \dfrac{5}{5}$
${a_y} = 1\dfrac{m}{{{s^2}}}$
According to the second equation of motion (in the y direction),
${S_y} = {u_y}t + \dfrac{1}{2}{a_y}{t^2}$
Here, there is no initial velocity, so,
${S_y} = \dfrac{1}{2}{a_y}{t^2}$
On putting the required values,
\[{S_y} = \dfrac{1}{2} \times 1 \times {4^2}\]
\[{S_y} = \dfrac{1}{2} \times 8\]
${S_y} = 4m$
On resolving the x and y vector, we get,
${S^2} = S_x^2 + S_y^2$${S^2} = {6^2} + {8^2}$
On putting the value of ${S_x}$ and ${S_y}$,
${S^2} = {6^2} + {8^2}$
${S^2} = 36 + 64$
${S^2} = 100$
On taking square root on both the sides,
$S = 10m$
So, the displacement of the block from the point where the force started acting is $S = 10m$.

Note: The components of a force represent the combined vertical and horizontal forces that combine to make the resultant force. It can also be said that any force acting in this universe has both the vertical as well as the horizontal component of force.