Question

A block of mass $10kg$ is moving horizontally with a speed of $1.5m{s}^{-1}$ on a smooth plane. If a constant force $10N$ acts on it perpendicular to initial velocity then the displacement of the block from the point of application of the force at the end of $4$ seconds is:

• A. $5m$
• B. $20m$
• C. $12m$
• D. $10m$

Answer 1

The correct option is D $10m$

Path of the block may be taken as parabolic.
In the X direction constant velocity

$S_x=u_x+\frac{1}{2}{a}_x{t}^2=1.5\times 4+0=6m{S_y=a}_y+\frac{1}{2}{a}_y{t}^2=0\times t+\frac{1}{2}[6\times 1]=8m\therefore displacement=\sqrt{S_x^2+S_y^2}=\sqrt{8^2+6^2}=10m$