A box of mass m , kg is placed on the rear si | vedclass

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

Question

(D) To find the acceleration of the box with respect to the truck,we analyze the forces acting on the box in the non-inertial frame of reference of th

Vedclass · Accepted Answer

It could be any value

Vedclass · Answer

(D) To find the acceleration of the box with respect to the truck,we analyze the forces acting on the box in the non-inertial frame of reference of the truck.$1$. The truck is accelerating forward with $a = 4 \, m/s^2$. Therefore,a pseudo force $F_p = ma$ acts on the box in the backward direction.$2$. The frictional force $f$ acts in the forward direction to oppose the tendency of motion relative to the truck. The maximum static friction is $f_{max} = \mu N = \mu mg$.$3$. For the box to remain at rest with respect to the truck,the pseudo force must be balanced by the frictional force:$ma = \mu mg$$4$. Dividing both sides by $m$ (assuming $m 
eq 0$):$a = \mu g$$5$. Substituting the given values:$4 = 0.4 	imes 10$$4 = 4$Since the equation $a = \mu g$ is independent of the mass $m$,the condition for the box to remain stationary with respect to the truck is satisfied for any value of $m$,provided the static friction is sufficient to balance the pseudo force.

$A$ 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 $[g = 10 \, m/s^2]$.

Similar Questions

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 ............ $N$.

$A$ block is kept on a frictionless inclined surface with an angle of inclination '$\alpha$'. The incline is given a horizontal acceleration '$a$' to keep the block stationary relative to the incline. Then '$a$' is equal to:

$A$ 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 $g = 10\,ms^{-2}$)

Find the apparent weight of a body of mass $1.0 \,kg$ falling with an acceleration of $10 \,ms^{-2}$. $\left(g \approx 10 \,ms^{-2}\right)$

Vedclass Test Series

Exam Paper Generator

Online Exam Module