Question

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

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

Answer 1

Answer: (D)

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

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