If a man of mass $50 \,kg$ is in a lift moving down with an acceleration equal to acceleration due to gravity, then the apparent weight of the man is

$A$ box of mass $m \, kg$ is placed on the rear side of an open truck accelerating at $4 \, m/s^2$. The coefficient of friction between the box and the surface below it is $0.4$. The net acceleration of the box with respect to the truck is zero. The value of $m$ is $[g = 10 \, m/s^2]$.

An object dropped in a stationary lift takes time $t_1$ to reach the floor. It takes time $t_2$ when the lift is moving up with constant acceleration. Then

$A$ 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

$A$ wooden wedge of mass $M$ and inclination angle $\alpha$ rests on a smooth floor. $A$ block of mass $m$ is kept on the wedge. $A$ force $F$ is applied on the wedge as shown in the figure such that the block remains stationary with respect to the wedge. The magnitude of the force $F$ is:

The weight of a man in a lift moving with the same acceleration upwards is $608 \ N$,while the weight of the same man in the lift moving downwards with the same acceleration is $368 \ N$. His normal weight is ............ $N$.