Suppose you are navigating a spacecraft far from other objects. The mass of the spacecraft is 2.0 104 kg (about 20 tons). The rocket engines are shut off, and you're coasting along with a constant velocity of ‹ 0, 28, 0 › km/s. As you pass the location ‹ 4, 4, 0 › km you briefly fire side thruster rockets, so that your spacecraft experiences a net force of ‹ 7.0 105, 0, 0 › N for 21 s. The ejected gases have a mass that is small compared to the mass of the spacecraft. You then continue coasting with the rocket engines turned off. Where are you an hour later? (Think about what approximations or simplifying assumptions you made in your analysis. Also think about the choice of system: what are the surroundings that exert external forces on your system?)

   r = <2.640 10⁶, 1.01 10⁸, 0> m

Explanation:

For this exercise we are going to solve it for each direction separately,

we locate a fixed reference frame in space at the height of the rocket, such that the position of the rocket is

  r₀ = <4, 4, 0> 10³ m

X axis

the initial velocity of the ship on this axis is v₀ₓ = 0, when it passes through the point x₀ = 4 km it ignites the rockets, experiencing a force of Fₓ = 7.0 10⁵ N for 21 s and the rockets turn off

They ask where it is after one hour t = 1 h = 3600 s

Let's apply Newton's second law

        Fₓ = m aₓ

        aₓ = Fₓ / m

        aₓ = 7.0 10⁵/2 10⁴

        aₓ = 3.5 10¹ m / s

Let's use kinematics to find the distance

for the first t₁ = 21 s the movement is accelerated

        x₁ = x₀ + v₀ t₁ + ½ aₓ t₁²

        x₁ = x₀ + ½ aₓ t₁²

        x₁ = 4000 + ½ 35 21² = 4000 + 7717,.5

        x₁ = 11717.5 m

this instant has a speed of

        vₓ = v₀ₓ + aₓ t

        vₓ = aₓ t ₁

        vₓ = 35  21

        vₓ = 735 m / s

the rest of the time there is no acceleration so it is uniform motion at this speed

        t₂ = 3600 - 21

        t₂ = 3576 s

         vₓ = x₂ / t₂

         x₂ = vₓ t₂

         x₂ = 735  3576

         x₂ = 2,628 10⁶ m

the total distance traveled in this direction is

         x_total = x₁ + x₂

         x_total = 11717.5 + 2.628 10⁶

         x_total = 2,640 10⁶ m

Y axis

on this axis it is in the initial position of y₀ = 4000 m, with an initial velocity of = 28 10³ m / s and there is no force on this axis F_{y} = 0

The movement in this axis is uniform,

       v_{y} = y / t

       y = v_{y} t

       y = 28 10³/3600

       y = 1.01 10⁸ m

the total distance is

       y_total = y₀ + y

       y_total = 4000 + 1.01 10⁸

       y_total = 1.01 10⁸ m

Z axis

the initial position is z₀ = 0, with an initial velocity of v₀ = 0 and in this axis there is no force F_{z} = 0

the movement is uniform

         z = 0

the final position of the rocket after 1 h is

      r = <2.640 10⁶, 1.01 10⁸, 0> m

the answer is kevlar

dec 23, 2016 - a load of bricks weighing 680 n a distance of 10 m across a parking lot. if if the worker pushes the cart with a constant force of 209 n, what work= force x distance = 220 n x 10 m = 2200 n.m = 2200 j i don't know where the person got it from, but the initial part, work= force x distance me.