An object with mass 60 kg moved in outer space. When it was at location < 13, -19, -3 > its speed was 3.5 m/s. A single constant force < 220, 320, -120 > N acted on the object while the object moved from location < 13, -19, -3 > m to location < 18, -11, -8 > m. Then a different single constant force < 150, 230, 220 > N acted on the object while the object moved from location < 18, -11, -8 > m to location < 22, -17, -3 > m. What is the speed of the object at this final location?
final speed = m/s

The speed of the object at its' final location is; 38 m/sWhat is work energy theorem?For the first force, we are given;Force; F₁ = 220i + 320j - 120kInitial Position; r₁ = 13i - 19j - 3kFinal Position; r₂ = 18i - 11j - 8kThus; Displacement; Δr = r₂ - r₁Δr = 18i - 11j - 8k - (13i - 19j - 3k)Δr = 5i + 8j - 5kFrom 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 - 3.5²1420 = v₂² - 12.25Solving gives v₂ = 37.85 m/sFor the second force, we are given;Force; F₂ = 150i + 230j - 220kInitial Position; r₁ = 18i - 11j - 8kFinal Position; r₂ = 22i - 17j - 3kThus; Displacement; Δr = r₂ - r₁Δr = 22i - 17j - 3k - (18i - 11j - 8k)Δr = 4i - 6j + 5kFrom 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 - 37.85²10.67 = v₂² - 1,432.6225Solving gives v₂ = 38 m/sRead more about Work Energy theorem at; https://brainly.com/question/14468674

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