A 200 kg load is raised using a 110 m long lever as shown...
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
Correct answer is A
ε= 80%, L=200× 10 = 2000N, dL=10m
dE= 110-10 = 100m, E=?
ε = workdoneontheloadworkdonebytheeffort × 100%
⇒ work done on load = 2000 ×10 =20,000 j
⇒ work done by effort = E × 100 = 100E
⇒ 80 = 20,000100E × 100%
⇒ 80 = 20,000E
⇒ E = 20,00080
⇒ E = 250N
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