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Question Number 68714 by Learner-123 last updated on 15/Sep/19

Answered by mr W last updated on 15/Sep/19

acceleration of object C=((1.2)/2)=0.6m/s^2   force in cable=(1/2)×375×((10+0.6)/(10))=198.75 N  power input=((198.75×0.6)/(0.85))=140.3 watt

$${acceleration}\:{of}\:{object}\:{C}=\frac{\mathrm{1}.\mathrm{2}}{\mathrm{2}}=\mathrm{0}.\mathrm{6}{m}/{s}^{\mathrm{2}} \\ $$$${force}\:{in}\:{cable}=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{375}×\frac{\mathrm{10}+\mathrm{0}.\mathrm{6}}{\mathrm{10}}=\mathrm{198}.\mathrm{75}\:{N} \\ $$$${power}\:{input}=\frac{\mathrm{198}.\mathrm{75}×\mathrm{0}.\mathrm{6}}{\mathrm{0}.\mathrm{85}}=\mathrm{140}.\mathrm{3}\:{watt} \\ $$

Commented by Learner-123 last updated on 15/Sep/19

please elaborate “force in cable”.

$${please}\:{elaborate}\:``{force}\:{in}\:{cable}''. \\ $$

Commented by mr W last updated on 16/Sep/19

weight of object = 375 N  mass of object =((375)/g)=37.5 kg  let force in cable =T  2T−mg=ma  ⇒T=((m(g+a))/2)=((375(10+0.6))/(2×10))=198.75 N

$${weight}\:{of}\:{object}\:=\:\mathrm{375}\:{N} \\ $$$${mass}\:{of}\:{object}\:=\frac{\mathrm{375}}{{g}}=\mathrm{37}.\mathrm{5}\:{kg} \\ $$$${let}\:{force}\:{in}\:{cable}\:={T} \\ $$$$\mathrm{2}{T}−{mg}={ma} \\ $$$$\Rightarrow{T}=\frac{{m}\left({g}+{a}\right)}{\mathrm{2}}=\frac{\mathrm{375}\left(\mathrm{10}+\mathrm{0}.\mathrm{6}\right)}{\mathrm{2}×\mathrm{10}}=\mathrm{198}.\mathrm{75}\:{N} \\ $$

Commented by Learner-123 last updated on 18/Sep/19

thanks sir.

$${thanks}\:{sir}. \\ $$

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