Question

A cart of mass $\mathrm{M}$ has a block of mass m attached to it, the coefficient of friction between the block and the cart is $\mathrm{\mu }$. What is the maximum acceleration of the cart so that the block m does not fall $\left(\mathrm{a}\right)$

• A. $\frac{\mathrm{\mu }}{\mathrm{g}}$
• B. $\mathrm{\mu g}$
• C. $\frac{\mathrm{g}}{\mathrm{\mu }}$
• D. $\frac{\mathrm{M\mu g}}{\mathrm{m}}$

Answer 1

The correct option is B

$\mathrm{\mu g}$

Step 1: Given data

Mass of cart $=\mathrm{M}$

Block $=\mathrm{m}$

Coefficient of friction $=\mathrm{\mu }$

Step 2. Concept to be used

Newton's second law of motion states the relationship between an object's mass and the amount of force needed to accelerate it. Newton's second law is often stated as F=ma, which means the force $\left(\mathrm{F}\right)$ acting on an object is equal to the mass $\left(\mathrm{m}\right)$ of an object times its acceleration $\left(\mathrm{a}\right)$.

So,

$\mathrm{F}=\mathrm{ma}$

Step 3. Determine the maximum acceleration of the cart so that the block m does not fall.

Consider the acceleration of the cart as $\mathrm{\alpha }$.

The free-body diagram of the block with respect to the cart is,

The block is at rest with respect to the cart. So,

$\mathrm{ma}=\mathrm{f}$

Where $\mathrm{f}$ is the total force, $\mathrm{m}$ is the mass of the system and $$a$$ is the net acceleration of the system.

$\mathrm{ma}=\mathrm{\mu N}$

Here, $\mathrm{N}$ is the normal force acting.

$\mathrm{\alpha }=\frac{\mathrm{\mu N}}{\mathrm{m}}$

Along the normal direction,

$\mathrm{N}=\mathrm{mg}$

So,

$\mathrm{\alpha }=\frac{\mathrm{\mu }\cancel{\mathrm{m}}\mathrm{g}}{\cancel{\mathrm{m}}}$

$\mathrm{\alpha }=\mathrm{\mu g}$

If the acceleration of the cart is less than $\mathrm{\alpha }=\mathrm{\mu g}$ then the block will not fall.

Hence, option $\mathrm{B}$ is the correct answer.