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)

10 m

Answer 2

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 $4 s$ $x =v t=1.5 \times 4=6\, m$ $\therefore s =\sqrt{x^{2}+y^{2}}=\sqrt{6^{2}+8^{2}}$ $=\sqrt{36+64}=10 \,m$