Question

a block of mass 10 kg is moving horizontally with a speed of 1.5 m/s on a smooth plane of a constant vertical force 10N acts on it. the displacement of the block from the point of application of force at the end of 4s is ?

Answer 1

Answer:

10m

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 meter per second square

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.