Physics, asked by maysmithjames5543, 10 months ago

A lift is moving downwards with an acceleration of 1.8m/s .what is apparent weight realised by a man of mass 50kg

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Answered by Anonymous
7

 \underline{ \boxed{ \bold{ \huge{ \purple{Answer}}}}} \\  \\  \star \sf \:  \bold{Given} \\  \\  \mapsto \sf \: mass  = 50 \: kg \\  \\  \mapsto \sf \: acceleration =  1.8 \:  \frac{m}{ {s}^{2} } \: downward  \\  \\  \mapsto \sf \: gravitational \: acceleration = 10 \:  \frac{m}{ {s}^{2} }  \: downward \\  \\  \star \sf \:  \bold{To \: Find} \\  \\  \mapsto \sf \: apparent \: weight of \: a \: man \\  \\  \star \sf \:  \bold{Formula} \\  \\  \mapsto \sf \: here \: real \: acc. \: is \: in \: downward \: direction \\  \sf \: so \: pseudo \: force \: acts \: in \: upward \: direction. \\  \\  \mapsto \sf \: \boxed{W{ \tiny{app}} = mg - F{ \tiny{pseudo}}} \\  \\   \star \sf \:  \bold{Calculation} \\  \\  \mapsto \sf \: W{ \tiny{app}} =  (50 \times 10) - (50 \times 1.8) \\  \\  \mapsto \sf \: W{ \tiny{app}} = 500 - 90 = 410 \: N \\  \\  \therefore \boxed{ \sf{ \bold{ \red{W{ \tiny{app}} = 410 \: N}}}}

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