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Difficulty: EasyWork, Energy and Power

A porter carries a suitcase of mass 5 kg5\text{ kg} along a horizontal platform for a distance of 10 m10\text{ m} at a constant speed, and then lifts it vertically upward through a height of 2 m2\text{ m} onto a shelf. Taking g=10 m s−2g = 10\text{ m s}^{-2}, what is the total work done by the porter on the suitcase?

  1. 100 J100\text{ J}Answer
  2. B
    600 J600\text{ J}
  3. C
    500 J500\text{ J}
  4. D
    10 J10\text{ J}

Answer

100 J100\text{ J}
During horizontal motion at constant speed, the upward force exerted by the porter is perpendicular to the horizontal displacement, so no work is performed horizontally (Whorizontal=Fscos⁡90∘=0 JW_{\text{horizontal}} = F s \cos 90^\circ = 0\text{ J}). During vertical lifting, the force acts in the direction of displacement, yielding Wvertical=mgh=5 kg×10 m s−2×2 m=100 JW_{\text{vertical}} = mgh = 5\text{ kg} \times 10\text{ m s}^{-2} \times 2\text{ m} = 100\text{ J}. The total work done is therefore 100 J100\text{ J}.

Step-by-Step Solution

1
Calculate the work done during horizontal motion
Whorizontal=0 JW_{\text{horizontal}} = 0\text{ J}
The supporting force acts vertically upward at an angle of 90∘90^\circ to the horizontal displacement, giving W=Fscos⁡90∘=0 JW = F s \cos 90^\circ = 0\text{ J}.
2
Calculate the work done in lifting the suitcase vertically
Wvertical=mgh=5×10×2=100 JW_{\text{vertical}} = mgh = 5 \times 10 \times 2 = 100\text{ J}
Work done against gravity equals the increase in gravitational potential energy.
3
Sum the work done in both stages
Wtotal=0+100=100 JW_{\text{total}} = 0 + 100 = 100\text{ J}
Total work done is the scalar sum of work performed along each segment of motion.

Key Concept

Work done by a constant force depends on the direction of displacement (W=Fscos⁡θW = F s \cos \theta). Perpendicular forces do no work.
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Topics

Work, Energy and Power

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