Question
A block of mass 10 kg rests on a horizontal floor. The acceleration due to gravity is 9.81 m/s2. The coefficient of static friction between the floor and the block is 0.2. A horizontal force of 10 N is applied on the block as shown in the figure. The magnitude of force of friction (in N) on the block is _______.
Answer :
10
W =mg= 10•9.81 = 98.1 N
Normal force, N = W = 98.1 N
Now, Maximum value of static friction force = µsN = 0.2•98.1 = 19.62 N
Applied force is less than static friction force, so friction force will be same as applied force. Hence, Friction force = 10N
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