A porter lifts a 10 kg luggage from the ground and places it on his head, 2.5 m above the ground. Calculate the work done by him on the luggage. (Take g = 10 ms⁻².)
250 J
The work done (W) is calculated using the formula:
\(W = Fd\cos\theta\)
where:
- \(F\) is the force applied.
- \(d\) is the displacement.
- \(\theta\) is the angle between the force and displacement.
In this case, the force is the weight of the luggage, which is given by:
\(F = mg\)
where:
- \(m\) is the mass (10 kg).
- \(g\) is the acceleration due to gravity (10 ms⁻²).
Therefore,
\(F = 10 \text{ kg} \times 10 \text{ ms}^{-2} = 100 \text{ N}\)
The displacement \(d\) is 2.5 m (the height the luggage is lifted).
The angle \(\theta\) between the force (acting vertically upwards) and displacement (also vertically upwards) is 0°. Therefore, \(\cos\theta = \cos(0°) = 1\).
Substituting these values into the work done formula:
\(W = 100 \text{ N} \times 2.5 \text{ m} \times 1 = 250 \text{ J}\)
Therefore, the work done by the porter on the luggage is 250 J.
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