Question

A block of mass $10kg$ is moving horizontally with a speed of $1.5{ms}^{-1}$ on a smooth plane. If a constant vertical force $10N$ acts on it, the displacement of the block from the point of application of the force at the end of $4s$ is

• A. $5m$
• B. $20m$
• C. $12m$
• D. $10m$
• E. $18m$

Answer 1

Answer: (D)

$10m$

$S_{horizontal}=ut=1.5\times 4=6m$
$S_{vertical}=\frac{1}{2}a{t}^2=\frac{1}{2}\frac{F}{m}{t}^2$
$=\frac{1}{2}\times \frac{10}{10}\times {\left(4\right)}^2=8m$
$S_{net}=\sqrt{{\left(6\right)}^2+{\left(8\right)}^2}$
$=\sqrt{36+64}=\sqrt{100}=10m$