(B) When a body of mass $m$ is in a lift moving upwards with an acceleration $a$,the forces acting on the body are the normal reaction $R$ (upwards) a

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(B) When a body of mass $m$ is in a lift moving upwards with an acceleration $a$,the forces acting on the body are the normal reaction $R$ (upwards) and the weight $mg$ (downwards).According to Newton's second law of motion,the net force is equal to the product of mass and acceleration:$F_{net} = ma$Here,the net force is $R - mg$ because the lift is accelerating upwards.Therefore,the equation of motion is:$R - mg = ma$Rearranging the terms to solve for $R$:$R = mg + ma$

Class 11 Physics Newton's Laws of Motion and Friction Apparent weight and Pseudo Force

Easy

If a body of mass $m$ is carried by a lift moving with an upward acceleration $a$,then the forces acting on the body are:
$(i)$ The reaction $R$ on the floor of the lift acting upwards.
$(ii)$ The weight $mg$ of the body acting vertically downwards.
The equation of motion will be given by:

A
$R = mg - ma$
B
$R = mg + ma$
C
$R = ma - mg$
D
$R = mg \times ma$

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Newton's Laws of Motion and Friction

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Apparent weight and Pseudo Force

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If a body of mass $m$ is carried by a lift moving with an upward acceleration $a$,then the forces acting on the body are:$(i)$ The reaction $R$ on the floor of the lift acting upwards.$(ii)$ The weight $mg$ of the body acting vertically downwards.The equation of motion will be given by:

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You stand on a spring scale on the floor of an elevator. Of the following,the scale shows the highest reading when the elevator

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$A$ car is moving on a circular horizontal track of radius $10 \, m$ with a constant speed of $10 \, m/s$. $A$ plumb bob is suspended from the roof of the car by a light rigid rod of length $1.00 \, m$. The angle made by the rod with the vertical is ........ $^o$.

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If a body of mass $m$ is carried by a lift moving with an upward acceleration $a$,then the forces acting on the body are:
$(i)$ The reaction $R$ on the floor of the lift acting upwards.
$(ii)$ The weight $mg$ of the body acting vertically downwards.
The equation of motion will be given by: