A 200 kg load is raised using a 110 m long lever as shown in the diagram above. The load is 10m from the pivot P. If the efficiency of the the lever is 80%, find the effort E required to lift the load.
[Take g = 10ms-2]
-
250
- 300
- 450
- 200
The correct answer is: A
Explanation
ε= 80%, L=200× 10 = 2000N, \(d_L\)=10m
\(d_E\)= 110-10 = 100m, E=?
ε = \(\frac{work done on the load}{work done by the effort}\) × 100%
⇒ work done on load = 2000 ×10 =20,000 j
⇒ work done by effort = E × 100 = 100E
⇒ 80 = \(\frac{20,000}{100E}\) × 100%
⇒ 80 = \(\frac{20,000}{E}\)
⇒ E = \(\frac{20,000}{80}\)
⇒ E = 250N
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