A1400-kg rocket has a net propulsion force of 20 kn (kilonewtons). over a short time period, it uniformly speeds up from an initial velocity of 35 m/s to a final velocity of 50 m/s. assume that the mass of the rocket is constant during this time period and that the net force is along the direction of motion. what is the net work done on the rocket in kilojoules (kj)?

  W = 8.92 10² kJ

Explanation:

For this exercise they give us the strength, we must calculate the distance traveled, for this we need the rocket acceleration let's use Newton's second law

        F = m a

        a = F / m

        a = 20 103/1400

        a = 14.29 m/s²

With kinematics we can find the distance traveled

      ² = v₀² + 2 a x

      x = ( ²-v₀²) / 2 a

      x = (50² -35²) / 2 14.29

      x = 1275 / 28.58

      x = 44.61 m

Let's calculate the work

      W = F.d

The bold is vector; as indicated by the force is in the direction of movement the scalar product is reduced to the ordinary product

     W = F d

     W = 20 10³ 44.61

     W = 8.92 10⁵ J

     W = 8.92 10² kJ

answer: electrons

15.3 cos 65 degrees trig