A body weighing 50 kg(on earth) is standing on a floor of the lift. Find the weight when --
1) the lift moves up with uniform velocity.
2)lift moves down with an acceleration of -4.9m/s²
3)lift moves up with an acceleration of 4.9m/s²
4)lift begans to fall freely under gravity.

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Solution

The actual weight of the body=mg=50 x 9.8=490N
(1)When lift is moving uniformly in upward direction,then acceleration of body=0
apparent weight =actual weight=490 N
(2)Lift moves down with an acceleration of -4.9m/s²
then apparent weight=490-50x4.9=245 N
(3)When lift moves up with an acceleration of 4.9m/s²
apparent weight=490+50x4.9=735N
(4)In free fall of the lift under gravity
Apparent weight=mg-mg=0

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Similar questions

List - I	List-II
(a) Lift moves up with uniform acceleration a	(d) mg
(b) Lift moves down with uniform acceleration a	(e) m(g $+$ a)
(c) Lift moves down with uniform velocity	(f) m(g-a)

m

\begin{matrix} List-I & List-II Lift moves & reading of machine a) up with uniform acceleration ' a' & e) m g b) down with uniform acceleration ' a' & f) m (g + a) c) down with uniform velocity & g) m (g - a) d) freely falling lift & h) 0 \end{matrix}

W_{1}

W_{2}

m

List I	List II
a) Lift is moving up with acceleration a	e) Apparent weight is greater than true weight
b) Lift is moving down with acceleration a	f) Apparent weight is zero
c) Lift is moving with uniformly velocity	g) Apparent weight is equal to true weight
d) Lift is freely falling	h) Apparent weight is less than true weight

1.2 m / s^{2} i i

g = 9.8 m / s^{2}

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