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Question Number 81124 by TawaTawa last updated on 09/Feb/20

Answered by Rio Michael last updated on 10/Feb/20

b) (i) Energy output = work done in lifting cement to the height B = mgsinθ×5 = 50sin60×5 = 216.5J ii) Energy input = Fd = 500 × 5 = 2500J (iii) Efficiency = ((work output)/(work inpu)) × 100% = ((216.5)/(2500)) × 100 = 8.66% iv) potential energy is converted to kinetic energy which is converted to sound and heat energy.

$$\left.\mathrm{b}\right)\:\left(\mathrm{i}\right)\:\mathrm{Energy}\:\mathrm{output}\:=\:\mathrm{work}\:\mathrm{done}\:\mathrm{in}\: \\ $$$$\mathrm{lifting}\:\mathrm{cement}\:\mathrm{to}\:\mathrm{the}\:\mathrm{height}\:\mathrm{B}\:=\:{mgsin}\theta×\mathrm{5} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{50}{sin}\mathrm{60}×\mathrm{5}\:=\:\mathrm{216}.\mathrm{5}{J} \\ $$$$\left.\mathrm{ii}\right)\:\mathrm{Energy}\:\mathrm{input}\:=\:{Fd}\:=\:\mathrm{500}\:×\:\mathrm{5}\:=\:\mathrm{2500}{J} \\ $$$$\left(\mathrm{iii}\right)\:\mathrm{Efficiency}\:=\:\frac{\mathrm{work}\:\mathrm{output}}{\mathrm{work}\:\mathrm{inpu}}\:×\:\mathrm{100\%} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{216}.\mathrm{5}}{\mathrm{2500}}\:×\:\mathrm{100}\:=\:\mathrm{8}.\mathrm{66\%} \\ $$$$\left.\mathrm{iv}\right)\:\mathrm{potential}\:\mathrm{energy}\:\mathrm{is}\:\mathrm{converted}\:\mathrm{to}\:\mathrm{kinetic}\:\mathrm{energy} \\ $$$$\mathrm{which}\:\mathrm{is}\:\mathrm{converted}\:\mathrm{to}\:\mathrm{sound}\:\mathrm{and}\:\mathrm{heat}\:\mathrm{energy}. \\ $$

Commented by TawaTawa last updated on 10/Feb/20

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

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