A block of mass 2kg is placed on the floor. The coefficient of static friction is $μ_{s} = 0.4$ . If a force of 2.8 N is applied on the block parallel to the floor, then the force of friction between the block and the floor (taking g=10m/sec2) is how much ?

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Solution

Step1: Given data
The mass of the block $m = 2 k g$ .
The coefficient of friction $μ_{s} = 0.4$ .
Force applied on the block $F_{n e t} = 2.8 N$
Step2: Formula used
The force of friction, $f = μ N [w h e r e μ = c o e f f i c i e n t, N = n o r m a l f o r c e]$
Step3: Calculating the force of friction between the block and the floor
Analyzing the given data in the figure.
We know that the value of N with the help of the figure i.e. $N = m g$
Putting this value of N in the force of friction formula, $f = μ m g$
We will put the given data,
$f = 0.4 \times 2 \times 10 f = 8 N$
As we can see that the above value of friction i.e. 8 N is more than the net force, $f > f_{n e t}$
Since the applied force is less than the force of friction, the friction adjusts its value and the new value becomes 2.8N so that block remains in equilibrium.
Thus, $f = 2.8 N$
Hence, the force of friction between the block and the floor is 2.8 N

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