Question

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

Answer 1

10 meters

Explanation:

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/s^2`
s(y) = `1/2 a(y) t^2 = 1/2 (1) (4^2) = 8 meters`

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