A lift is moving downwards with an accelerati | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The normal force $N$ acting on the body of mass $m$ inside the lift can be calculated using Newton's second law.For a lift moving downwards with a

Vedclass · Accepted Answer

Zero

Vedclass · Answer

(D) The normal force $N$ acting on the body of mass $m$ inside the lift can be calculated using Newton's second law.For a lift moving downwards with acceleration $a$,the equation of motion is $mg - N = ma$.Given that the lift is moving downwards with an acceleration equal to the acceleration due to gravity,we have $a = g$.Substituting $a = g$ into the equation: $mg - N = mg$.This simplifies to $N = mg - mg = 0$.The frictional force $f$ is given by $f = \mu N$.Since the normal force $N = 0$,the frictional resistance $f = \mu 	imes 0 = 0$.Therefore,the correct option is $D$.

$A$ lift is moving downwards with an acceleration equal to the acceleration due to gravity. $A$ body of mass $m$ kept on the floor of the lift is pulled horizontally. If the coefficient of friction is $\mu$,then the frictional resistance offered by the body is:

Similar Questions

Select the correct statement regarding pseudo force.

The ratio of the weight of a man in a stationary lift and when it is moving downward with uniform acceleration $a$ is $3 : 2$. The value of $a$ is ($g$ = acceleration due to gravity of the earth).

$A$ rough vertical board has an acceleration $a$ so that a $2 \ kg$ block pressing against it does not fall. The coefficient of friction between the block and the board should be

The ratio of the weight of a man in a stationary lift to his weight in a lift moving downwards with an acceleration '$a$' is $3:2$. Then '$a$' = . . . . . . .

$A$ mass is hanging on a spring balance which is kept in a lift. The lift ascends with an acceleration. The spring balance will show in its readings:

Vedclass Test Series

Exam Paper Generator

Online Exam Module