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:

$W_a=<-400,500,250>.<-1,4.2,-13.8>$

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

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

Answer 2

Work done is the energy required to displace an object. The amount of work that is been done by your friend is 1717.34 J.

What is work done?

Work done is the energy required to displace an object.

As we know that the position of the space shuttle is changing, while the velocity of the space shuttle also increases. therefore,

The Work done is the change in kinetic energy and also the work done in changing the position of the space shuttle,

Work = Change in Kinetic energy = work done by you + work done by a friend

$\rm W_{KE}=W_A +W_B$

$\dfrac{1}{2}m(v_1^2-v_o^2) =W_B + (F\cdot ds)$

As it is given that the mass of the satellite is 700 kg, and its velocity changes from 6 m/s to 6.18 m/s, while the position changes from <3.3, -1.6, 2.0> m to <2.3, 2.6, -11.8> m, and the force that is been applied during the entire time is  <-400, 500, 250> N. therefore,

$\dfrac{1}{2}\times 700 \times(6.18^2-6^2) \\=W_B - [ < -400, 500, 250 > \cdot ( < 2.3, 2.6, -11.8 > - < 3.3, -1.6, 2.0 > )]\\\\767.34 =W_B + ( < -400, 500, 250 > \cdot < -1, 4.2, -13.8 > )\\\\767.34 =W_B+ (400 + 2100-3450)\\\\ 767.34 = W_B -950\\\\ W_B = 1717.34\rm\ J$

Hence, the amount of work that is been done by your friend is 1717.34 J.

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