Question

A 40 kg slab rests on a frictionless floor. A 10 kg block rests on top of the slab. The static coefficient of friction between the block and the slab is 0.60 while the kinetic coefficient is 0.40. The 10 kg block is acted upon by horizontal force 100 N. $g = 9.8m/{s^2}$ , then the resulting acceleration of the slab will be

${\text{A}}{\text{. 0}}{\text{.98m/}}{{\text{s}}^2}$
${\text{B}}{\text{. 1}}{\text{.47m/}}{{\text{s}}^2}$
${\text{C}}{\text{. 1}}{\text{.52m/}}{{\text{s}}^2}$
${\text{D}}{\text{. 6}}{\text{.1m/}}{{\text{s}}^2}$

Answer 1

Hint: First analyse if the blocks move together or not, and for that find the acceleration of the total mass which is (40 + 10) kg then use the formula F = ma to find a by substituting the value of F and m and then solve further.

Formula used - $F = ma$ , ${f_m} = {\mu _s} \times N$ , $N = mg$ , $F = \mu mg$

Complete Step-by-Step solution:
Given that a 40 kg slab rests on a frictionless floor and a 10 kg block rests on top of the slab.
Also, the static coefficient of friction between the block and the slab is ${\mu _s} = 0.60$
And the kinetic coefficient is $\mu = 0.40$
Let the mass of block be ${m_1} = 10kg$ and mass of slab is $m = 40kg$
Now, let us first analyse that if these blocks move together.
For that we will be finding the acceleration of the total mass, which means mass of slab + mass of block, i.e., (40 + 10) kg.
Now using the formula $F = Ma$
or we can write- $a = \dfrac{F}{M}$
Now, given the value of F is 100 N in the question and total mass is 40 + 10 = 50 kg.
So, $a = \dfrac{{100}}{{50}} = 2m/{s^2}$
Now the maximum frictional force is ${f_m} = {\mu _s} \times N$
Also, $N = {m_1}g$ so the maximum frictional force will be –
${f_s} = {\mu _s} \times {m_1}g$
So, putting the values of friction coefficient = 0.60, mass of block is 10 kg and $g = 9.8m/{s^2}$ we get-
${f_s} = (0.60)(10)(9.8) = 58.8N$
Thus, we can see frictional force is smaller than the applied force and proves that the block and the slab do not move relative to each other.
Now, for the block we have $F = \mu {m_1}g$
Putting the values of the known terms, we get-
$F = 0.4 \times 10 \times 9.8 = 39.2N$
Now, the resulting acceleration of the slab will be, $a = \dfrac{F}{m} = \dfrac{{39.2}}{{40}} = 0.98m/{s^2}$
Hence, the resulting acceleration of the slab is $0.98m/{s^2}$ .
Therefore, the correct option is A.

Note – Whenever such types of questions appear then always write down the things given in the question then apply then find the acceleration of the total mass due to the application of a horizontal force of 100 N on the block of 10 kg. And then find the maximum frictional force on the block, then we observed that the frictional force is smaller than the applied force and it proves that the block and the slab are not moving relative to each other. So, we can write for the block that $F = \mu {m_1}g$ and then after finding the value of F, we can have the value of a by using the formula, $a = \dfrac{F}{m}$ .