Question

Evaluate the final kinetic energy of the supply spacecraft for the actual tractor beam force, F(x)=αx3+β .

Answer 1

Answer:

$\large \boxed{8 \times 10^{9} \text{ J}}$

Explanation:

$\begin{array}{rcl}KE_{2} - KE_{1}&=&\int_{x_{1}}^{ x_{2}}F(x)dx\\KE_{2}- 2.7 \times 10^{11}&=&\int_{0}^{ 7.5 \times 10^{4}}(-3.5 \times 10^{6})dx\\&=&-3.5 \times 10^{6}\int_{0}^{ 7.5 \times 10^{4}}dx\\&=&-3.5 \times 10^{6}(7.5 \times 10^{4})\\& = & -2.62 \times 10^{11}\\KE_{2} & = & 2.7 \times 10^{11} - 2.62 \times 10^{11}\\& = & 0.08 \times 10^{11}\\& = & \mathbf{8 \times 10^{9}} \textbf{ J}\\\end{array}$

$\text{The final kinetic energy of the supply spacecraft would have been \large \boxed{\mathbf{8 \times 10^{9}} \textbf{ J}} }$

Answer 2

Answer:

7.5 × 10^9 J.

Explanation:

So, we are given the following data or parameters or information for the proper solving of the question above;

=> ''actual force exerted by the tractor beam as a function of position is given by F(x) = ax + , where a = 6.1 x 10 N/mº and B = -4.1 x 10° N''

=> " Assume the supply spacecraft had an initial kinetic energy of KE = 2.7 x 10" J "

=> "that the tractor beam force is applied on the spacecraft over a distance of 7.5 x 10^11 m away.from its beginning position at = 0.0 m"

So, we can then solve it as;

KE2 – KE1 = F(x) dr.

KE2 - 2.7 x 10^11 J = - 3.5 × 10^6 × 7.5 x 10^4.

KE2 = 7.5 × 10^9 J.