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Difficulty: MediumWork, Energy, and Power

An industrial robotic manipulator in an assembly plant applies a steady vertical force of 400 N400\text{ N} to lift a structural bracket through a height of 3 meters3\text{ meters}. If the manipulator delivers a constant power output of 150 W150\text{ W} throughout the movement, how much time does the manipulator take to complete the lift?

  1. A
    3 seconds3\text{ seconds}
  2. 8 seconds8\text{ seconds}Answer
  3. C
    0.125 seconds0.125\text{ seconds}
  4. D
    50 seconds50\text{ seconds}

Answer

The robotic manipulator takes 8 seconds8\text{ seconds} to complete the lift.
The correct answer is found by first determining the mechanical work performed and then dividing by the power output. The work done lifting the bracket is W=F×d=400 N×3 m=1200 JW = F \times d = 400\text{ N} \times 3\text{ m} = 1{}200\text{ J}. Since power is the rate of work done (P=WtP = \frac{W}{t}), the time required is t=WP=1200 J150 W=8 secondst = \frac{W}{P} = \frac{1{}200\text{ J}}{150\text{ W}} = 8\text{ seconds}.

Step-by-Step Solution

1
Calculate the total mechanical work done to lift the bracket.
W=Force×Distance=400 N×3 m=1200 JW = \text{Force} \times \text{Distance} = 400\text{ N} \times 3\text{ m} = 1{}200\text{ J}
Work is defined as force multiplied by the distance over which that force is applied in the direction of motion.
2
Calculate the time elapsed using the power relationship.
t=WP=1200 J150 W=8 secondst = \frac{W}{P} = \frac{1{}200\text{ J}}{150\text{ W}} = 8\text{ seconds}
Power represents the rate of doing work (P=WtP = \frac{W}{t}), which rearranges to t=WPt = \frac{W}{P}.

Key Concept

Relationship between mechanical work, force, distance, and power
Estimated Time:1m 0s
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Topics

Work, Energy, and Power

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