Question

Outside the space shuttle, you and a friend pull on two ropes to dock a satellite whose mass is 700 kg. The satellite is initially at position <3.3, -1.6, 2.0> m and has a speed of 6 m/s. You exert a force <-400, 500, 250> N. When the satellite reaches the position <2.3, 2.6, -11.8> m, its speed is 6.18 m/s. How much work did your friend do?

Answer 1

Answer:

your friend did a work of 1717.34 J

Explanation:

The sum of the work made by your friend and you is equal the change in the kinetic energy, so:

$W_a+W_b=\frac{1}{2}mv_f^{2}-\frac{1}{2}mv_i^{2}$

Where $W_a$ is the work that you did, $W_b$ is the work that your friend did, m is the mass of the satellite and $v_f$ and $v_i$ are the final and initial speeds

Additionally the work that you did is equal to:

$W_a=F.s$

Where F and s are vectors. So, the vector s that said the displacement is calculated as:

s = <2.3, 2.6, -11.8> - <3.3, -1.6, 2.0>

s = <-1 , 4.2 , -13.8>

Therefore, $W_a$ is:

$W_a=.$

$W_a=(-400*-1)+(500*4.2)+(250*-13.8)\\W_a=400+2100-3450\\W_a=-950$

Then, replacing the values on initial equation and solving for $W_b$, we get:

$-950+W_b=\frac{1}{2}(700)(6.18)^{2}-\frac{1}{2}(700)(6)^{2}\\-950+W_b=767.24\\W_b=767.24+950\\W_b=1717.34J$

Finally, your friend did a work of 1717.34 J