An engine of one metric ton is going up an inclined plane $$1$$ in $$2$$ at the rate of $$36$$ $$km/hr$$. If the coefficient of friction is (1/$$√$$3), the power of the engine in watt is

Question

Infinity Learn · Accepted Answer

The situation is shown in fig,Power $$P=F$$×VGiven $$v=36km/hr=36$$×$$518m/sF=mgsin$$⁡θ$$+f=mgsin$$⁡θ$$+$$μmgcos⁡θ $$=mg($$$$sin$$⁡θ$$+$$μ$$cos$$⁡θ$$)$$ $$=1000$$×$$9$$⋅$$8(0$$⋅$$5+13$$×$$32)$$ $$($$∵$$sin$$⁡θ$$=12$$ and $$cos$$⁡θ$$=32)$$ $$=9800N$$∴$$P=9800$$×$$10=98k$$ watt.

An engine of one metric ton is going up an inclined plane 1 in 2 at the rate of 36 km/hr. If the coefficient of friction is (1/√3), the power of the engine in watt is

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