Question

A $40kg$ slab rests on a frictionless floor. A $10kg$ block rests on top of the slab. The coefficient of friction between the block and the slab is $0.40$. The $10kg$ block is acted upon by a horizontal force of $100N$. If $g=10m/s^2$, the resulting acceleration of the slab will be

• A. $1.0m/s^2$
• B. $1.47m/s^2$
• C. $1.52m/s^2$
• D. $6.1m/s^2$

Answer 1

The correct option is A $1.0m/s^2$

• Maximum value of friction between slab and block is $\mu N$ as block has no motion in vertical direction so $normal$ $reaction$ $N=mg=10g=10\times 10=100newton$
• Thus maximum friction is $0.4\times 100Newton=40Newton$
• This friction is the only horizontal force which act on the slab so acceleration of Slab = $\frac{friction}{40Kg}$= $1$ $m{s}^{-2}$
• Option A is correct.
• Note if we treat slab and block both as a system then common acceleration is $\frac{100N}{10Kg+40Kg}$ which is $2m{s}^{-2}$ but friction is only able to produce $1m{s}^{-2}$ so block will slip over slab.