At t = 0 a 100gm ball is thrown upwards with initial speed `v_(0) = 2` m/s in an open platform elevator which at that time is moving downward with v = 3 m/s and accelerating downward with an acceleration, `a = 3 m//s^(2)`​. A drag force with magnitude `F_"​DRAG"` = 0.8 N acts on the ball. Net force acting on the ball in the frame of the elevator just after it has been thrown is [`g =10 m//^(2)`​] :

Options

A. 1.2 N
B. 1.5 N
C. 1.8 N
D. 2.4 N

Answer

B

Explanation

Elevator is non inertial frame of reference, therefore we must apply pseudo force.
Elevator is accelerating downwards, therefore pseudo force must act upwards while mg and `F_"DRAG​"` act downwards.
`therefore F_"NET"` [downward] = mg -ma + `F_"DRAG"`
= 0.1 × 10 – 0.1 × 3 + 0.8 = 1.5 N.