"A 2-kg object is moving at 3 m/s. A 4-N force is applied in the direction of motion and then removed after the object has traveled an additional 5 m. What is the work done by this force? What is the velocity of the object after the force was applied?"

The velocity of the object after the force was applied is equal to 5.39 m/s.

Given the following data:

a. To find the work done by this force:

Mathematically, the work done by an object is given by the formula;

Substituting the values into the formula, we have;

Work done = 20 Joules

b. To determine the velocity of the object after the force was applied, we would apply the law of conservation of energy:

Kinetic energy after - Kinetic energy before = Work done.

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Explanation:

Given that,

Mass m = 2kg

Initial Velocity Vi = 3m/s

Force applied. = 4N

Distance cause by the force applied d = 5m

a. Work done by the force applied?

Work done by force applied can be determined by

W = F×d•Cosx

Where x is direction is the force and displacement

The force and displacement are in the same direction, then, x =0°

W = F×d•Cosx

W = 4×5Cos0

W = 20J.

b. Final velocity

The change in kinetic energy is equal to work done

∆K.E = W

½mVf² - ½mVi² = W

½mVf² = ½mVi² + W

½ × 2 Vf² = ½ × 2 × 3² + 20

Vf² = 9+20

Vf² = 29

Vf = √29

Vf = 5.39m/s

The final velocity is 5.39m/s

a) the event horizon, singularity, and the chute located between the two.