Question

A block of mass $10kg$ is moving horizontally with a speed of $1.5m{s}^{-1}$ 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 $4$ second is

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

Answer 1

Answer: (D)

Here, $m=10kg,F=10N$
The situation is as shown in the figure.

For motion along vertical direction
Acceleration along vertical direction is $a_y=\frac{F}{m}=\frac{10N}{10kg}=1ms{}^{-2}$
Distance travelled by the block in $4s$ in vertical direction is
$S_y=\frac{1}{2}{a}_y{t}^2=\frac{1}{2}\times \left(1m{s}^{-2}\right)(4s{)}^2=8m$
For motion along horizontal direction
Distance travelled by the block in $4s$ in horizontal direction is
$S_x=\left(1.5m{s}^{-1}\right)(4s)=6m$
The displacement of the block at the end of $4s$
$=\sqrt{S_x^2+S_y^2}=\sqrt{(8m{)}^2+(6m{)}^2}=10m$