A block of mass 5 kg is moving horizontally at a speed of 1.5 m per second a perpendicular force of 5 Newton acts on it for 4 second what will be the distance of the block from the point where the force started acting
Answers
Answered by
35
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 = 1ms2
s(y) = 12a(y)t2=12(1)(42)=8meters
Resolving the x and y vector, we note the Pythagorean Triple of 6, 8, 10. So the distance travelled is 10 meters.
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 = 1ms2
s(y) = 12a(y)t2=12(1)(42)=8meters
Resolving the x and y vector, we note the Pythagorean Triple of 6, 8, 10. So the distance travelled is 10 meters.
Similar questions