Question

A person weighing 60 kg is standing on a platform scale balance kept on the floor of an elevator
cab. The elevator cab is capable of motion up or down with either a uniform velocity or a constant
acceleration, the value of velocity as well as acceleration may be adjusted at any suitable value.
Newton's second law of motion can be applied for the motion of elevator cab.
Din which direction lift is moving if the acceleration of lift is in upward direction
(a)up
(b)down
(c)first up and then down
(dycan not say as direction of velocity not given
(1what will be observed weight of the person as recorded by the balance when elevator cab is
moving upward with a constant velocity of 2 m/s
(a)60 kg
6) 0 kg
(C)600 kg
(d)10 kg
(iiy what will be observed weight of the person as recorded by the balance when elevator cab is
moving upward with a constant acceleration i.e g4
(a)60 kg
(6)75 kg
(C)40 kg
(d)0 kg
jv) what will be observed weight of the person as recorded by the balance when elevator cab is
moving downward with a velocity of 20 m/s is being stopped and in the process a deceleration of
g/2 is acting on it
(a) 70 kg
(b)90 kg
(C)50 kg
(d)0 kg
The elevator cab starts falling downward freely due to some mechanical failure in its control
system, the observed weight is
(a)70 kg
(b)90 kg
(C)50 kg
(d)0 kg​

Answer 1

Answer:

A person is standing on a weighing machine placed on the floor of an elevator. The elevator starts going up with same acceleration, moves with uniform velocity for a while and finally decelerates to stop. The maximum and the maximum weights recorded are 72 kg and 60 kg.

hope this helps you ☺️

please mark me as brainliest.

Answer 2

(i) The answer is (d) can not say as the direction of velocity not given.

The acceleration of the lift alone can not tell the direction of the motion of the lift. We need to know the direction velocity of the lift. The acceleration could be positive or negative.

(ii) The answer is (a) 60kg

If the lift is moving upwards with a constant velocity of 2 m/s. The velocity is constant means there is no acceleration. Due to the absence of acceleration, a pseudo force on the objects inside the elevator will not be present and hence the weight of the objects will be the same as in an elevator at rest.

(iii) Given: mass of the person = 60kg

Acceleration of the elevator is g/4 in the upwards direction

To Find: Weight of the body observed by the weighing machine

Solution:

Since the elevator is moving upwards with a constant acceleration of g/4, the bodies inside the elevator will experience a pseudo force in the downwards direction. The pseudo force is given as F = ma, where m is the mass of the body and a is the acceleration

The net force acting in downwards direction = Gravitation force + pseudo force

F net = ma + mg

Fnet = m(a + g)

Fnet = 60 (g + g/4)

       = 60 x $\frac{5g}{4}$

       = 750 N

Therefore, the weight observed by the weighing machine is (b)75kg

(iv) The velocity of the elevator is downwards, so it is moving in the downward direction. If it is decelerating with g/2 then the acceleration must be opposite to the velocity. Hence, the acceleration of the lift is in the upwards direction. The pseudo force will act downwards.

Fnet = Gravitaional force + pseudo force

Fnet = m (g + a)

       = 60 (3g/2)

       = 900N

Therefore, the observed weight is (b)90kg

(v)If the elevator is freely falling, the acceleration is equal to g and is in the downwards direction. So, the pseudo force will act in the upward direction. Here the pseudo force will be equal to mg and in the opposite direction of the gravitational force.  Both of these forces will cancel out and there will be a sensation of weightlessness.

Therefore, the weighing scale will read (d)0kg.