Question Number 68714 by Learner-123 last updated on 15/Sep/19
![](https://www.tinkutara.com/question/9297.png)
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](https://www.tinkutara.com/question/Q68772.png)
$${acceleration}\:{of}\:{object}\:{C}=\frac{\mathrm{1}.\mathrm{2}}{\mathrm{2}}=\mathrm{0}.\mathrm{6}{m}/{s}^{\mathrm{2}} \\ $$$${force}\:{in}\:{cable}=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{375}×\frac{\mathrm{10}+\mathrm{0}.\mathrm{6}}{\mathrm{10}}=\mathrm{198}.\mathrm{75}\:{N} \\ $$$${power}\:{input}=\frac{\mathrm{198}.\mathrm{75}×\mathrm{0}.\mathrm{6}}{\mathrm{0}.\mathrm{85}}=\mathrm{140}.\mathrm{3}\:{watt} \\ $$
Commented by Learner-123 last updated on 15/Sep/19
![please elaborate “force in cable”.](https://www.tinkutara.com/question/Q68813.png)
$${please}\:{elaborate}\:“{force}\:{in}\:{cable}''. \\ $$
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](https://www.tinkutara.com/question/Q68919.png)
$${weight}\:{of}\:{object}\:=\:\mathrm{375}\:{N} \\ $$$${mass}\:{of}\:{object}\:=\frac{\mathrm{375}}{{g}}=\mathrm{37}.\mathrm{5}\:{kg} \\ $$$${let}\:{force}\:{in}\:{cable}\:={T} \\ $$$$\mathrm{2}{T}−{mg}={ma} \\ $$$$\Rightarrow{T}=\frac{{m}\left({g}+{a}\right)}{\mathrm{2}}=\frac{\mathrm{375}\left(\mathrm{10}+\mathrm{0}.\mathrm{6}\right)}{\mathrm{2}×\mathrm{10}}=\mathrm{198}.\mathrm{75}\:{N} \\ $$
Commented by Learner-123 last updated on 18/Sep/19
![thanks sir.](https://www.tinkutara.com/question/Q69040.png)
$${thanks}\:{sir}. \\ $$