Question

A block of mass $5kg$ is moving horizontally at a speed of 1.5 m/s. A perpendicular force of 5 N acts on it for 4 sec. What will be the distance of the block from the point where the force started acting

• A. 10 m
• B. 8 m
• C. 6 m
• D. 2 m

Answer 1

Answer: (A)

10 m

Answer 2

In the given problem force is working in a direction perpendicular to initial velocity. So the body will move under the effect of constant velocity in horizontal direction and under the effect of force in vertical direction.

$S_{x} = u_{x} \times t = 1.5 \times 4 = 6m$

$S_{y} = u_{y}t + \frac{1}{2}at^{2} = 0 + \frac{1}{2}(F/m)t^{2}$ $= \frac{1}{2}(5/5)(4)^{2}$ $= 8m$

$\therefore$ $S = \sqrt{S_{x}^{2} + S_{y}^{2}} = \sqrt{36 + 64} = \sqrt{100} = 10m$