Question

A block of mass 5kg is moving horizontally at a speed of 1.5m/s.A perpendicular force of 5N acts on it for 4 sec.what will be the displacement of the block from the point when the force start acting?answer is 10m

Answer 1

Assume initial velocity of 1.5 m/s is in the x-direction
Since there are no forces on it in this direction, there will be no acceleration. So distance = s(x) = 1.5 m/s (4 sec) = 6 meters
In the y-direction, F=5N and since m=5 kg, Newton's 2nd Law tells us acceleration a(y) = F/m = 5/5 = 1 N/kg = 1 m/sec^2
s(y) = 1/2a(y)t^2 = 1/2 1*4^2= 8 m

Resolving the x and y vector, we note the Pythagorean Triple of 6, 8, 10. So the distance travelled is 10 meters.

Answer 2

Ans-10m ,option-A

by phytagogar triplet 8,6,10