Question

A block of mass $10$ kg is moving horizontally with a speed of $1.5m{s}^{-1}$ on a smooth plane. If a constant vertical force $10$ $N$ acts on it, 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. $18m$
• D. $10m$

Answer 1

The correct option is C $18m$

Given velocity of the block$v=1.5m/s$ and force applied $F=10N$

Now we know $F=ma$

$a=1m/s^2$

Also $a=\frac{v-u}{t}$ here t=4 s (given)

$-u=at-v$

$u=2.5m/s$

Now from equation of motion we get,

$s=ut+\frac{1}{2}a{t}^2$

$s=\left(2.5\right)\times 4+8=18m$