A block of mass 2kg is placed on the floor. The coefficient of static friction is . 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 ?
Step1: Given data
The mass of the block .
The coefficient of friction .
Force applied on the block
Step2: Formula used
The force of friction,
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.
Putting this value of N in the force of friction formula,
We will put the given data,
As we can see that the above value of friction i.e. 8 N is more than the net force,
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,
Hence, the force of friction between the block and the floor is 2.8 N