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Question
A coolie carries a load of 30 kgf through a distance of 500 m in 5 minutes while another coolie B carries the same load through the same distance in 10 minutes. Compare the
(i) work done, and
(ii) power developed. (Take: g = 10md−2)
Solution
Given : F = 30 kgf = 30 × g N = 30 × 10 N = 300 N, d = 500 m.
For coolie A, work done W1 = F × d = 300 × 500 = 150,000 J
Time t1 = 5 minute = 5 × 60s = 300s
∴ Power developed P1 =`"W"_1/"t"_1=(150,000)/300` = 500 Js−1
For coolie B, work done W2 = F × d = 3000 × 500 = 150,000 J
Time t2 = 10 minutes = 10 × 60s = 600s
∴ Power developed P2 =`"W"_2/"t"_2=(150,000)/600` = 250 Js−1
(i) `"W"_1/"W"_2=(15,000)/(15,000)=1/1.`
(ii) `"P"_1/"P"_2=500/250=1/2.`
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