A 0.52 kg object is at rest. A 3.45 N force to the right acts on the object during a time interval of 1.55 s.A 0.52 kg object is at rest. A 3.45 N...

We assume that under the influence of the forces applied the object travels horizontally and there are no frictional forces to be overcome for this movement.

We know force is equal to mass multiplied by acceleration, that is:

f = m*a

or a = f/m

Where f = force, m = mass and a = acceleration

Let f1 be the initial force applied to the right and a1 the resultant acceleration. Then:

a1 = f1/m = 3.45/0.52

as the force is applied for 1,55 s (t1) the acceleration will continue for that much time.

Therefore final velocity v1 at the end of 1.55 s is given by formula:

v1 = a1*t1 = (3.45/0.52)*1.55 = 10.2836 m/s^2 to the right (approximately)

To find the velocity after the force o5 4.41 N is applied to the left for 3.38 s.

a2 = f2/m = -4.41/0.52

and velocity after after 3.38 seconds

v2 = v1 + a2*t2 = 10.2836 + (-4.41/0.52)*3.38 = 10.2836 + (28.665)

= -18.3814 m/s^2 to the right

or 18.3814 m/s^2 to the left