A 60, kg man stands on a spring scale in a li | vedclass

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

Question

(C) The apparent weight $W'$ of a person in a lift is given by $W' = m(g + a)$,where $a$ is the acceleration of the lift.If the lift is moving upwards

Vedclass · Accepted Answer

The lift,while in constant motion upwards,is stopped suddenly.

Vedclass · Answer

(C) The apparent weight $W'$ of a person in a lift is given by $W' = m(g + a)$,where $a$ is the acceleration of the lift.If the lift is moving upwards with a constant velocity,the acceleration $a = 0$,so the reading is $m 	imes g = 60\, kg$.When the lift suddenly stops while moving upwards,it experiences a downward acceleration (retardation),so $a$ becomes negative. The apparent weight becomes $W' = m(g - a)$.Since the reading decreases from $60\, kg$ to $50\, kg$,it indicates that the lift experienced a downward acceleration (deceleration) while moving upwards.Therefore,we conclude that the lift was moving upwards with constant speed and was suddenly stopped.

$A$ $60\, kg$ man stands on a spring scale in a lift. At some instant,he finds the scale reading has changed from $60\, kg$ to $50\, kg$ for a while and then comes back to the original mark. What should we conclude?

Similar Questions

The free surface of oil in a tanker,at rest,is horizontal. If the tanker starts accelerating,the free surface will be tilted by an angle $\theta$. If the acceleration is $a \ m/s^2$,what will be the slope of the free surface?

$A$ cork is submerged in water by a spring attached to the bottom of a bowl. When the bowl is kept in an elevator moving with an acceleration $a$ downwards, the length of the spring:

$A$ man is standing in a lift which goes up and comes down with the same constant acceleration. If the ratio of the apparent weights in the two cases is $2 : 1$,then the acceleration of the lift is ......... $m/s^2$.

$A$ block of mass $m$ is placed on a smooth wedge of inclination $\theta$. The whole system is accelerated horizontally so that the block does not slip on the wedge. The force exerted by the wedge on the block ($g$ is acceleration due to gravity) will be

$A$ person in an elevator accelerating upwards with an acceleration of $2\,ms^{-2}$,tosses a coin vertically upwards with a speed of $20\,ms^{-1}$. After how much time will the coin fall back into his hand? (Take $g = 10\,ms^{-2}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module