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Pulley

A lift moving...

Question

A lift moving upwards with a velocity 10 m/s is
stopped by uniform retardation in 5 s. The apparent
weight of a man of mass 60 kg before stopping will
be (take g = 10 m/s2)
(1) 72 kg wt (2) 60 kg wt
(3) 48 kg wt (4) 50 kg wt

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Solution

$w h e n a m a n g o e s u p w i t h u = 10 m s^{- 1}, h e s t o p s i n 5 s e c o n d . T h e r e f o r e a c c o r d i n g t o f i r s t e q n o f m o t i o n, 0 = 10 - a \times 5 o r a = 2 m s^{- 1}, i s r e t a r d i n g a c c e l e r a t i o n a c t i n g o n t h e m a n . S o n e t r e t a r d a t i o n = (g + a) = (10 + 2) = 12 m s^{- 1} T h e r e f o r e a p p a r e n t w e i g h t o f t h e m a n = m (g + a) = \frac{w}{g} (g + a) = \frac{60}{10} \times 12 = 72 K g w t .$

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