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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

$$\mathrm{b})\left(\mathrm{i}\right)\mathrm{Energy}\mathrm{output}=\mathrm{work}\mathrm{done}\mathrm{in}$$$$\mathrm{lifting}\mathrm{cement}\mathrm{to}\mathrm{the}\mathrm{height}\mathrm{B}=mgsin\theta \times 5$$$$=50sin60\times 5=216.5J$$$$\mathrm{ii})\mathrm{Energy}\mathrm{input}=Fd=500\times 5=2500J$$$$\left(\mathrm{iii}\right)\mathrm{Efficiency}=\frac{\mathrm{work}\mathrm{output}}{\mathrm{work}\mathrm{inpu}}\times 100\mathrm{\%}$$$$=\frac{216.5}{2500}\times 100=8.66\mathrm{\%}$$$$\mathrm{iv})\mathrm{potential}\mathrm{energy}\mathrm{is}\mathrm{converted}\mathrm{to}\mathrm{kinetic}\mathrm{energy}$$$$\mathrm{which}\mathrm{is}\mathrm{converted}\mathrm{to}\mathrm{sound}\mathrm{and}\mathrm{heat}\mathrm{energy}.$$

Commented by TawaTawa last updated on 10/Feb/20

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

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