Question

A block of mass 5 kg is moving horizontally at a speed of 1.5 ${ms}^{-1}$. A vertically upward force 5 N acts on it for 4 seconds. 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

The correct option is D 10 m

Assume initial velocity of $1.5m/s$ is in the $x-$direction

Since there are no forces on it in this direction$,$ there will be no acceleration$.$ So distance $=s\left(x\right)=1.5m/s\left(4sec\right)=6m$

In the $y-$direction$,F=5N$ and since $m=5kg,$ Newton's $2nd$ Law tells us acceleration $a\left(y\right)=F/m=5/5=1N/kg=1m{s}^2$

$s\left(y\right)=12a\left(y\right)t_2=12\left(1\right)\left(42\right)$=$8m$

Resolving the $x$ and $y$ vector, we note the Pythagorean Triple of $6,8,10.$

So$,$ the distance traveled is $10m$

Hence,

option $\left(D\right)$ is correct answer.