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 a horizontal force of 100 N. If g = 9.8m/s2, then what will be the resulting acceleration of the slab?

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Solution

## Step1: Given dataGiven that a $40kg$ slab rests on a frictionless floor and a $10kg$ block rests on top of the slab.Also, the static coefficient of friction between the block and the slab is ${\mathrm{Î¼}}_{s}=0.60$Step2: Formula used$F=ma\left[whereF=force,m=mass,a=acceleration\right]$Step3: Calculating the accelerationLet us first analyze if these blocks move together.For that, we will be finding the acceleration of the total mass, which means the $massofslab+massofblock$, i.e., $\left(40+10\right)kg$.Now using the formula $F=ma$or we can write- $a=\frac{F}{m}$Now, given the value of $F=100N$ in the question and the total mass is $40+10=50kg$.So, $a=\frac{100}{50}=2m/{s}^{2}$Now the maximum frictional force is f${f}_{m}={\mathrm{Î¼}}_{s}Ã—N$Also, $N={m}_{1}g$ so the maximum frictional force will be : ${f}_{s}={\mathrm{Î¼}}_{s}Ã—{m}_{1}g$So, putting the values of friction coefficient ${\mathrm{Î¼}}_{s}=0.60$, mass of block is ${m}_{1}=10kg$ and $g=9.8m/{s}^{2}$we get- $fs=\left(0.60\right)\left(10\right)\left(9.8\right)=58.8N$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 $F=\mathrm{Î¼}{m}_{1}g$Putting the values of the known terms, we get- $F=0.4Ã—10Ã—9.8=39.2N$Now, the resulting acceleration of the slab will be, $a=\frac{F}{m}=\frac{39.2}{40}=0.98m/{s}^{2}$Hence, the resulting acceleration of the slab is $0.98m/{s}^{2}.$

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