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- Apparent Weight Of Body In A Lift MCQ - Practice Questions with Answers
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9 Questions around this concept.
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MEDIUMA person of mass M is standing on lift. If lift moves vertically upwards according to given v-t graph then find out the apparent weight of man at t = 1 sec. (g = 10m/s2)
A man weighs 80 Kg . He stands on a weighing machine in a lift which is moving upwards with a uniform acceleration of $5 \mathrm{~m} / \mathrm{s}^2$.what would be the reading on the machine ( $g=10 \mathrm{~m} / \mathrm{s}^2$ )
Three blocks with masses m, 2 m and 3 m are connected by strings, as shown in the figure. After an upward force F is applied on block m, the masses move upward at constant speed . What is the net force on the block of mass 2m?
(g is the acceleration due to gravity)
A mass of 50 kg is placed in a lift and it is moving upwards with an acceleration of The apparent weight of the body is
A man with a weight of 50 kg is there in a lift; the lift is moving upward with a constant velocity; The reaction of the man is
A body is there on a lift which is going downwards with an acceleration ; The body will
Concepts Covered - 1
If mass m is placed on a weighing machine which is placed in the lift.
Actual weight = mg
Apparent weight = Reaction force
R - Reaction force is given by reading of weighing machine.
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Lift is at Rest
$\begin{aligned} & V=0, a=0 \\ & \text { use } F_{\text {net }}=m \vec{a} \\ & R-m g=0 \\ & R=m g\end{aligned}$
Apparent weight = Actual weight
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Lift is moving upward or down with constant Velocity
$\begin{aligned} & v=\text { const }, a=0 \\ & R=m g\end{aligned}$
Apparent weight = Actual weight
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Lift is accelerating Upward with acceleration = a
$\begin{aligned} & v=\text { variable, } \\ & a<g \\ & R-m g=m a \\ & R=m(g+a)\end{aligned}$
Apparent weight > Actual weight
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Lift is moving up with a = g
$\begin{aligned} & R-m g=m g \\ & R=2 m g\end{aligned}$
Apparent weight = 2 (Actual weight)
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