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

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10 m

8 m

6 m

2 m

answer is A.

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Detailed Solution

$S_{Horizontal} = ut = 1 .5 \times 4 = 6 m S_{Vertical} = \frac{1}{2} {at}^{2} = \frac{1}{2} \frac{F}{m} t^{2} = \frac{1}{2} \times 1 \times 16 = 8 m S_{Net} = \sqrt{6^{2} + 8^{2}} = 10 m$

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