Question

A block of mass $5 \,kg$ is moving horizontally at a speed of $ 1.5\,m{{s}^{-1}} $ . A vertically upward force $5 N$ acts on it for $4 s$. What will be the distance of the block from the point where the force starts acting?

• A. 2 m
• B. 6 m
• C. 8 m
• D. 10 m

Answer 1

Answer: (D)

Upward acceleration $=\frac{5}{5}=1 \,m / s ^{2}$
Upward distance covered in $4 s$
$y =\frac{1}{2} a t^{2}$
$=\frac{1}{2} \times 1 \times(4)^{2}=8 \,m$
Horizontal distance covered in $4 s$
$x =v t=1.5 \times 4=6\, m $
$\therefore s =\sqrt{x^{2}+y^{2}}=\sqrt{6^{2}+8^{2}} $
$=\sqrt{36+64}=10 \,m$