Question

A \[40\text{ }kg\] slab rests on a frictionless floor. A \[10\text{ }kg\] 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 \[10\text{ }kg\] block is acted upon by horizontal force 100 N. If \[g=9.8m/{{s}^{2}}\] then the resulting acceleration of the slab will be.

A. \[0.98m{{s}^{-2}}\]
B. \[1.47m{{s}^{-2}}\]
C. \[1.52m{{s}^{-2}}\]
D. \[6.1m{{s}^{-2}}\]

Answer 1

Hint: To solve the acceleration of the blocks (block and slab) moving we will first find the acceleration of the total (block and slab) moving by dividing the horizontal force by the total mass (block and slab) of the blocks in form of:
\[a\text{ }=\text{ }\dfrac{F}{m}\]
After that we will find the frictional force of the (block and slab) using the formula:
\[~F\text{ }=\text{ }\mu mg\]
where F is the force applied, u is the friction coefficient, m is the mass of the block and g is the gravity of \[9.8m{{s}^{-2}}\].

Complete step by step answer:
The free body diagram for the block and slab when in motion during both static and kinetic friction scenarios are:

First we assume that the block and slab move together as one with the total mass of the block and slab as sum of the mass of block 1 and slab i.e.
\[\Rightarrow \left( 10+40 \right)kg\]
Total mass of the block and slab combination is \[50kg\].
Now, the block of 10 kg mass is pulled with a force of 100N making the acceleration of the block family as:
\[a\text{ }=\text{ }\dfrac{F}{m}\]
Now to check if both the blocks moves or not, for that we will find the frictional force exerted by the block of mass \[10kg\] when pulled with a force of \[100N\], placing the values in the formula
\[~F\text{ }=\text{ }\mu mg\]
\[~F\text{ }=\text{ }0.6\times 10\times 9.8\]
\[~F\text{ }=\text{ }58.8N\]
As we can see that the force exerted by the friction on the block is 60N which is less than the force applied to pull the block, we will ignore the block of mass 10kg and will focus on the slab of 40kg and similarly like before we will use the formula:
\[~F\text{ }=\text{ }\mu mg\]
\[~F\text{ }=\text{ }0.4\times 10\times 9.8\]
\[~F\text{ }=\text{ }39.2N\]
The force exerted on the slab with a frictional force of \[0.40\] is \[39.2N\].
Now placing the force as \[39.2N\] and the mass as \[40kg\], we get the acceleration of the slab as:
\[a\text{ }=\text{ }\dfrac{F}{m}\]
\[a\text{ }=\text{ }\dfrac{39.2}{40}\]
\[a=0.98m{{s}^{-2}}\]
Therefore, the resulting acceleration on the block is given as \[a=0.98m{{s}^{-2}}\].

Note:
The coefficient of friction is divided into two parts static and kinematic, the static friction is the friction applied when the block is not moving and the kinetic friction is the friction applied when the object is moving with the friction applied on the block both static and kinetic friction can be applied together as well for the condition when block is both moving and not moving.