Question

A block of mass 5 kg is moving horizontally at a speed of 1.5 m/s. A perpendicular force of 5 N acts on it for 4 sec. What will be the shortest distance of the block from the point where the force started acting.

• A. 10 m
• B. 8 m
• C. 6 m
• D. 2 m

Answer 1

The correct option is A 10 m

Assume initial velocity of 1.5 m/s is in the x-direction

Since there are no forces on it in this direction, there will be no acceleration.

So, distance $S_x=1.5\times 4=6m$

In the y-direction, $F=5N$ and $m=5kg$

Acceleration in y=direction,

$a_y=\frac{F}{m}=\frac{5}{5}=1m/s^2$

$S_y=\frac{1}{2}{a}_y{t}^2=\frac{1}{2}\times 1\times 4^2=8m$

Resolving the x and y vector we get,

$S^2=S_x^2+S_y^2$

$S^2=6^2+8^2$

$S=\sqrt{36+64}$

$S=\sqrt{100}$

$S=10m$