A box of mass m kg is placed on the rear side of an open...

A box of mass $m \mathrm{~kg}$ is placed on the rear side of an open truck accelerating at $4 \mathrm{~ms}^{-2}$. 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 $\left[\right.$ Given $\left.g=10 \mathrm{~ms}^{-2}\right]$ (a) $4 \mathrm{~kg}$ (b) $8 \mathrm{~kg}$ (c) $9.78 \mathrm{~kg}$ (d) It could be any value

A box of mass $m \mathrm{~kg}$ is placed on the rear side of an open truck accelerating at $4 \mathrm{~ms}^{-2}$. 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 $\left[\right.$ Given $\left.g=10 \mathrm{~ms}^{-2}\right]$
(a) $4 \mathrm{~kg}$
(b) $8 \mathrm{~kg}$
(c) $9.78 \mathrm{~kg}$
(d) It could be any value

Key Concepts

Newton's Second Law

This fundamental law of mechanics states that the net force acting on an object is equal to the product of its mass and its acceleration. It is used to set up the relationship between the forces acting on the box and its acceleration, allowing us to understand how the frictional force can balance the inertial force due to acceleration.

Static Friction

Static friction is the force that resists the initiation of sliding motion between two surfaces in contact. It has a maximum value determined by the coefficient of static friction multiplied by the normal force. In problems like this, static friction is crucial as it prevents the box from slipping despite the truck’s acceleration.

Non-Inertial Reference Frame

Analyzing motion from a non-inertial (accelerating) reference frame requires the inclusion of pseudo forces to account for the observed acceleration. In this context, considering the truck as a non-inertial frame helps explain how the static friction force must counterbalance an effective inertial force acting on the box.

Limiting Condition for No Relative Motion (Maximum Friction Condition)

When the force required to prevent relative motion equals the maximum static friction available, the system is at a limiting case. This concept shows that under such circumstances, the condition for no sliding becomes independent of the mass of the box, leading to situations where any mass can satisfy the equilibrium condition provided the forces balance exactly.

Transcript

00:01 In this question we have given that the rear side of a truck is open and a box of mass 40 kage is placed 5 meter away from the open end and we have given the coefficient of friction rate is 0 .15 now it is told that on a straight road the truck starts from the rest and accelerate with 2 meter per second square then we have to find the distance traveled by the truck when the box falls from this truck we have given that this is a truck there and a block is placed in this truck and as the truck is moving in this direction with an acceleration a and this distance is given to us that is this is the open end of the truck that is as is equal to this is 5 meter and we have given that mass of the block that is m is equal to 40 kg and the coefficient of friction that is 0 .15 so we can say that when there is a friction force active on this block let this is equal to f here and there is a force acting which is equals to f in this direction that is equal to mass into acceleration so we can first of i am calculating the force acting on this block due to the acceleration so we can say that f is equal to this is m into a so we can say this will comes out to be 40 into acceleration that is equal to 2 meter per second square so we can say this is a 40 into 2 so so we can say this is equal to 80 newton.

01:35 And we can say that force of the friction that is f is equal to this is mu into mg and this is 0 .15 into 10.

01:43 So we can say this is equal to 60 newton.

01:47 So we can say that the net force acting on the block is that is equal to f minus small f and this is equal to mass into this is acceleration of the block.

02:11 So we can say this will come out to be that is 20 upon this is this is the acceleration of block...

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