Question

In a pulley system, for lifting a weight of 250 kgf through a height of 30 m, velocity ratio of 3 is used. If x effort is required in downward direction so that work done by effort is 810 J. Find the value of x.

A) 8 N
B) 9 N
C) 10 N
D) 5 N

PLEASE DO NOT SPAM AND EXPLAIN WITH SOLVING. BEST ANSWER = BRAINLIEST.

Answer 1

$\large \rm {\underbrace{\underline{Elucidation:-}}}$

$\cal \red {\maltese{\underline{\underline{Provided\: that:}}}}$

$\to \sf {Weight=25kgf}$

$\to \sf {Displacement\: of\: load(d_{L})=30m}$

$\to \sf {Velocity\: ratio(VR)=3}$

$\to \sf {Work\: done\: by\: effort(w_{E})=810J}$

$\cal \blue {\maltese{\underline{\underline{To\: determine:}}}}$

$\to \sf {Force\: applied\: on\: effort(F_{E})=?}$

$\cal \pink {\maltese{\underline{\underline{We\: know:}}}}$

$\sf {Work\: done\: by\: effort=Force\: applied\: on\: effort×Displacement\: of\: effort}$

$\cal {\boxed{\underline{w_{E}=F_{E}×d_{E}}}}$

➻To use this formula ,we need to calculate displacement of the effort.

$\cal \purple {\maltese{\underline{\underline{Using \: the\: formula:}}}}$

$\sf {Displacement\: of\: effort=displacement\: of\: load× velocity\: ratio}$

$\cal {\boxed{d_{E}=d_{L}× VR}}$

$\hookrightarrow \sf{d_{E}=30m× 3}$

$\hookrightarrow \sf \green {\underline{d_{E}=90m}}$

➻As we know the displacement of the effort, we can find out the force applied on the effort.

$\hookrightarrow \sf {w_{E}=F_{E}×d_{E}}$

$\hookrightarrow \sf {810J=F_{E}×90m}$

$\large \hookrightarrow \sf {F_{E}=\frac{810}{90}}$

$\large \hookrightarrow \sf {F_{E}=\frac{81{\cancel{0}}}{9{\cancel{0}}}}$

$\large \hookrightarrow \sf {F_{E}=\frac{81}{9}}$

$\large \hookrightarrow \sf {F_{E}=\frac{\cancel{81}^{9}}{\cancel{9}}}$

$\implies \cal \green {\boxed{\boxed{F_{E}=9N}}}$

$\cal \orange {\maltese{\underline{\underline{Henceforth,}}}}$

➻The force applied by the effort 'x' is "9 Newtons" respectively.

➻"Option-B" is the correct answer.

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