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)

Answer 1

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.`