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

Options

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

Answer

D

Explanation

Here , m=10 kg , F= 10 N
The situation is as shown in the figure .
Acceleration along vertical direction
`a_y=F/m ="10 N"/"10 kg" = 1 m s^(-2)`
Distance travelled by the block in 4 s in vertical direction is
`s_y=1/2a_yt^2=1/2xx(1 m s^(-2) )(4 s)^2`= 8 m
Distance travelled by the block in 4 s in horizontal direction is
`s_x=(1.5 m s^(-1))(4 s)`=6 m
The displacement of the block at the end of 4 s is
`=sqrt(s_x^2+s_y^2)=sqrt((8 m)^2 + (6 m)^2)`= 10 m