# Question-68714

Question Number 68714 by Learner-123 last updated on 15/Sep/19
Answered by mr W last updated on 15/Sep/19
$${acceleration}\:{of}\:{object}\:{C}=\frac{\mathrm{1}.\mathrm{2}}{\mathrm{2}}=\mathrm{0}.\mathrm{6}{m}/{s}^{\mathrm{2}} \\$$$${force}\:{in}\:{cable}=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{375}×\frac{\mathrm{10}+\mathrm{0}.\mathrm{6}}{\mathrm{10}}=\mathrm{198}.\mathrm{75}\:{N} \\$$$${power}\:{input}=\frac{\mathrm{198}.\mathrm{75}×\mathrm{0}.\mathrm{6}}{\mathrm{0}.\mathrm{85}}=\mathrm{140}.\mathrm{3}\:{watt} \\$$
Commented by Learner-123 last updated on 15/Sep/19
$${please}\:{elaborate}\:“{force}\:{in}\:{cable}''. \\$$
Commented by mr W last updated on 16/Sep/19
$${weight}\:{of}\:{object}\:=\:\mathrm{375}\:{N} \\$$$${mass}\:{of}\:{object}\:=\frac{\mathrm{375}}{{g}}=\mathrm{37}.\mathrm{5}\:{kg} \\$$$${let}\:{force}\:{in}\:{cable}\:={T} \\$$$$\mathrm{2}{T}−{mg}={ma} \\$$$$\Rightarrow{T}=\frac{{m}\left({g}+{a}\right)}{\mathrm{2}}=\frac{\mathrm{375}\left(\mathrm{10}+\mathrm{0}.\mathrm{6}\right)}{\mathrm{2}×\mathrm{10}}=\mathrm{198}.\mathrm{75}\:{N} \\$$
Commented by Learner-123 last updated on 18/Sep/19
$${thanks}\:{sir}. \\$$