An object with mass 60 kg moved in outer space. When it was at location \(\langle 13, -19, -3 \rangle\), its speed was 3.5 m/s.
A single constant force \(\langle 220, 320, -120 \rangle\) N acted on the object while the object moved from location \(\langle 13, -19, -3 \rangle\) m to location \(\langle 18, -11, -8 \rangle\) m.
Then a different single constant force \(\langle 150, 230, 220 \rangle\) N acted on the object while the object moved from location \(\langle 18, -11, -8 \rangle\) m to location \(\langle 22, -17, -3 \rangle\) m.
What is the speed of the object at this final location?
Final speed = ______ m/s
Answer (3)
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
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
The final speed of the object after moving under the influence of the two forces is approximately 4.99 m/s. This is determined using the work-energy theorem, which equates the work done by the forces to the change in kinetic energy. The calculations involve determining the work done by each force and solving for the respective speeds at the final positions.
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