Question

Outside the space shuttle, you and a friend pull on two ropes to dock a satellite whose mass is 800 kg. The satellite is initially at position < 3.0, -1.6, 3.0 > m and has a speed of 6 m/s. You exert a force < -400, 400, 250 > N. When the satellite reaches the position < 1.7, 2.6, 4.8 > m, its speed is 7.11 m/s. How much work did your friend do?

Answer 1

Solution:

Let us go to the basics first.

Let
r1 = initial position,
r2 = final position,
i,j,k = unit vectors along x,y,z axes respectively
F1 = force by me,
W1 = work done by me,
W2 = work done by my friend

r1 = (3 i - 1.6 j + 3 k)m
r2 = (1.7 i + 2.6 j + 4.8 k) m
F1 = (-400 i + 400 j + 250 k)N
W1 = F1.(r2 - r1)
Or W1 = (-400 i + 400 j + 250 k).{(1.7 i + 2.6 j + 4.8 k) - (3 i - 1.6 j + 3 k)}
Or W1 = (-400 i + 400 j + 250 k).(1.7 i + 2.6 j + 4.8 k - 3 i + 1.6 j - 3 k)
Or W1 = (-400 i + 400 j + 250 k).(-1.3 i + 4.2 j +1.8 k)
Or W1 = -400 * (-1.3) + 400 * 4.2 + 250 *1.8
Or W1 = 2650 J

Mass m = 800 kg
Initial speed u = 6 m/s
Initial kinetic energy K1 = 1/2 mu^2
Or K1 = 1/2 * 800 * 6^2
Or K1 = 14400 J

Final speed v = 7.11 m/s
Final kinetic energy K2 = 1/2 mv^2
Or K2 = 1/2 * 800 * 7.11^2
Or K2 = 20220.84 J

Total work done = change in kinetic energy
=> W1 + W2 = K2 - K1
=>2650 J + W2 = 20220.84 J - 14400 J
=> W2 = 3170.84 J (Answer)

Thanks!!!