an object of mass 2.5 kg moving horizontally in a direction due north at an initial speed of 4ms?

Question

is acted on by an applied force of 0.10 N for 20S.the force is directed horizontally in direction due east,calculate:
a)velocity of the object after applied force is removed.
b)the increase of K.E of the object.
c)the distance moved by the object eastwards.

Answer 1

The initial speed is irrelevant to to the applied force.

The velocity vector E is Ft/m = (0.10 N)(20 s)/(2.5 kg) = 0.80 m/s
The velocity is thus 0.80 m/s E + 4.0 m/s N

The change in KE is m/2*(vN^2 + vE^2) - m/2*vN^2 = mvE^2/2 =
(2.5 kg)*(0.80 m/s)^2/2 = 0.80 J

The distance travelled E is d = at^2/2 = Ft^2/(2m) =
(0.10 N)(20 s)^2/(2*2.5 kg) = 8.0 m