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]
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
