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Sagot :
Sigma F.dS = total work done = change in kinetic energy
(220, 320, -120).(18-13,-11+19,-8+3) +(150, 230, 220).(22-18,-17+11,-3+8)= 1/2 *60*(V^2- 3.5^2)
220*5+320*8+ -120*-5 + 150*4 + 230* 6 +220* -5= ..
simplify his
(220, 320, -120).(18-13,-11+19,-8+3) +(150, 230, 220).(22-18,-17+11,-3+8)= 1/2 *60*(V^2- 3.5^2)
220*5+320*8+ -120*-5 + 150*4 + 230* 6 +220* -5= ..
simplify his
The speed of the object at its' final location is; 38 m/s
What is work energy theorem?
For the first force, we are given;
Force; F₁ = 220i + 320j - 120k
Initial Position; r₁ = 13i - 19j - 3k
Final Position; r₂ = 18i - 11j - 8k
Thus; Displacement; Δr = r₂ - r₁
Δr = 18i - 11j - 8k - (13i - 19j - 3k)
Δr = 5i + 8j - 5k
From work energy theorem, we know that;
F₁ * Δr = ¹/₂m(v₂² - v₁²)
We are given v₁ = 2.5 m/s and m = 60 kg. Thus;
(220i + 320j - 120k) × (5i + 8j - 5k) = ¹/₂ * 60(v₂² - 3.5²)
4260/30 = v₂² - 3.5²
1420 = v₂² - 12.25
Solving gives v₂ = 37.85 m/s
For the second force, we are given;
Force; F₂ = 150i + 230j - 220k
Initial Position; r₁ = 18i - 11j - 8k
Final Position; r₂ = 22i - 17j - 3k
Thus; Displacement; Δr = r₂ - r₁
Δr = 22i - 17j - 3k - (18i - 11j - 8k)
Δr = 4i - 6j + 5k
From work energy theorem, we know that;
F₂ * Δr = ¹/₂m(v₂² - v₁²)
Now, v₁ = 37.85 m/s and m = 60 kg. Thus;
(150i + 230j + 220k) × (4i - 6j + 5k) = ¹/₂ * 60(v₂² - 37.85²)
320/30 = v₂² - 37.85²
10.67 = v₂² - 1,432.6225
Solving gives v₂ = 38 m/s
Read more about Work Energy theorem at; https://brainly.com/question/14468674
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