A person of mass M kg is standing on a lift. If the lift moves vertically upwards according to given v-t graph then find out the weight of the man at the following instants :(g=10m/ $s^{2}$ )
(i) t = 1 second
(ii) t= 8 seconds
(iii) t= 12 seconds

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Solution

Since, at time $1 sec$ the acceleration will be $10 m / s^{2}$ .

The weight is given as,

$W = m (g + a)$

$= M (10 + 10)$

$= 20 M$

Since, at time $8 sec$ the acceleration will be $0 m / s^{2}$ .

The weight is given as,

$W = m (g + a)$

$= M (10 + 0)$

$= 10 M$

Since, at time $12 sec$ the acceleration will be $- 10 m / s^{2}$ .

The weight is given as,

$W = m (g + a)$

$= M (10 - 10)$

$= 0$

Thus, the weight of man at $1 sec$ , $8 sec$ and $12 sec$ is $20 M$ , $10 M$ and $0$ respectively.

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A person of mass M kg is standing on a lift. If the lift moves vertically upwards according to given v-t graph then find out the weight of the man at the following instants :(g=10m/s2)(i) t = 1 second(ii) t= 8 seconds (iii) t= 12 seconds

A person of mass M kg is standing on a lift. If the lift moves vertically upwards according to given v-t graph then find out the weight of the man at the following instants :(g=10m/ $s^{2}$ )
(i) t = 1 second
(ii) t= 8 seconds
(iii) t= 12 seconds