Question

Galactic Alliance Junior Mission Officer (GAJMO) Bundit Nermalloy is predicting the kinetic energy of a supply spacecraft, which is being moved in one dimension in the tractor beam of the ship named the Jadarian-Ruby, to ensure that the supply spacecraft doesn’t damage the spaceport to which it is being delivered. GAJMO Nermalloy has been instructed to deliver the supply spacecraft with a kinetic energy less than 1010J (where 1J=1N⋅m). GAJMO Nermalloy knows that the change in kinetic energy of an object moving in one dimension is equal to the net work performed on it, where net work is the integral of the component of net force in the direction of motion with respect to the position of the of the object. That is: KE2−KE1=∫x2x1F(x)dx. The net force exerted by the tractor beam is supposed to be constant, F0=−3.5×106N, but due to improper maintenance of the Jadarian-Ruby, the actual force exerted by the tractor beam as a function of position x is given by F(x)=αx3+β, where α=6.1×10−9N/m3 and β=−4.1×106N. Assume the supply spacecraft had an initial kinetic energy of KE1=2.7×1011J and that the tractor beam force is applied on the spacecraft over a distance of 7.5×104m away.from its beginning position at x=0.0m.

Answer 1

Answer:

the ship's energy is greater than this and the crew member does not meet the requirement

Explanation:

In this exercise to calculate kinetic energy or final ship speed in the supply hangar let's use the relationship

                W =∫ F dx = ΔK

                 

Let's replace

             

          ∫ (α x³ + β) dx = ΔK

         α x⁴ / 4 + β x = ΔK

           

Let's look for the maximum distance for which the variation of the energy percent is 10¹⁰ J

         x (α x³ + β) = $K_{f}$ - K₀

          $K_{f}$  = K₀ + x (α x³ + β)

Assuming that the low limit is x = 0, measured from the cargo hangar

     

Let's calculate

        $K_{f}$  = 2.7 10¹¹ + 7.5 10⁴ (6.1 10⁻⁹ (7.5 10⁴) 3 -4.1 10⁶)

        Kf = 2.7 10¹¹ + 7.5 10⁴ (2.57 10⁶ - 4.1 10⁶)

        Kf = 2.7 10¹¹ - 1.1475 10¹¹

        Kf = 1.55 10¹¹ J

In the problem it indicates that the maximum energy must be 10¹⁰ J, so the ship's energy is greater than this and the crew member does not meet the requirement

We evaluate the kinetic energy if the System is well calibrated

                W = x F₀ = $K_{f}$ –K₀

                $K_{f}$ = K₀ + x F₀

We calculate

              $K_{f}$ = 2.7 10¹¹ -7.5 10⁴ 3.5 10⁶

               $K_{f}$ = (2.7 -2.625) 10¹¹

              $K_{f}$ = 7.5 10⁹ J

Answer 2

The value of final kinetic energy of the spacecraft is  $1.075 \times 10^{10} \;\rm J$.

Given data:

The value of F is, $F(x)= \alpha x^{3}+\beta$.

The constant beam force is, $F_{0}=-3.5 \times 10^{6} \;\rm N$.

The value of constants are, $\alpha = 6.1 \times 10^{-9} \;\rm N \;\;\;\rm and \;\;\;\;\rm \beta = -4.1 \times 10^{6}\;\rm N$.

The initial kinetic energy is, $KE_{1}=2.7 \times 10^{11} \;\rm J$.

And distance is, $x = 7.5 \times 10^{4} \;\rm m$

In the given problem, it is required to obtain the value of final kinetic energy. For this, use the relation of work- energy as,

$W= \int {F} \, dx= \Delta KE$

here, F is the force exerted by the tractor beam and dx is the position function.

Solving as,

$\int {(\alpha x^{3}+\beta)} \, dx=\Delta KE\\\\\dfrac{\alpha x^{4}}{4}+ \beta x=\Delta KE$

Let's look for the maximum distance for which the variation of the energy percent is  $10^{10} \;\rm J$.

$\dfrac{\alpha x^{4}}{4}+ \beta x= KE_{2}-KE_{1}\\\\\dfrac{6.1 \times 10^{-9} \times (7.5 \times 10^{4})^{4}}{4}+ (-4.1 \times 10^{6} \times 7.5 \times 10^{4})= KE_{2}-(2.7 \times 10^{11})\\\\KE_{2} = 1.075 \times 10^{10} \;\rm J$

Thus, we can conclude that the value of final kinetic energy of the spacecraft is  $1.075 \times 10^{10} \;\rm J$.

Learn more abut the work-energy theorem here:

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