Outside the space shuttle, you and a friend pull on two ropes to dock a satellite whose mass is 1500 kg. The satellite is initially at position < 3.2, -1, 2.3 > m and has a speed of 3 m/s. You exert a force < -500, 470, -300 > N. When the satellite reaches the position < 6.2, 3.5, 3.8 > m, its speed is 3.06 m/s. How much work did your friend do?
2007-06-29 03:13:43 · 3 answers · asked by Ashley 2 in Science & Mathematics ➔ Physics
The two of you did enough work to change the satellite's KE from 0.5*1500*3^2 to 0.5*1500*3.06^2 or 272.7 J. Since we don't know the satellite's initial and final velocity vectors or how long you applied your force, we can't say by how much your and your friend's individual forces aided or opposed the satellite's KE.
2007-06-29 03:19:36 · answer #1 · answered by kirchwey 7 · 0⤊ 1⤋
Total work done
= 0.5* 1500 * ( v +u) (v-u)
=0.5* 1500 * 6.6 * 0.6
= 2970 J
Work done by you
in the X direction = 500 x 3 =1500 J
in the Y direction = 470 x 4.5 =2115
in the Z direction = - 300 x 1.5 = - 450 J
Total work done by you = 3165
Work done by your friend is 2970 – 3165 = -195 J.
2007-06-29 11:32:49 · answer #2 · answered by Pearlsawme 7 · 0⤊ 0⤋
calculate what the guy above me stated, then multiply by distance to get work
work = force (energy) times distance
qouted from the guy above me: Enough to change its KE from 0.5*1500*3^2 to 0.5*1500*3.06^2.
2007-06-29 10:24:10 · answer #3 · answered by Flaming Pope 4 · 0⤊ 1⤋