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Skip to questionJEE Main · Physics · Laws of Motion
A 2 kg block remains at rest on a rough horizontal surface. The coefficient of static friction is 0.5 and g = 10 m/s². A horizontal force of 6 N is applied. What is the static friction force?
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The block remains at rest. What must be true horizontally?
Congrats! You built the answer yourself.
You completed every hint without a wrong choice.The block is at rest, so the net horizontal force must be zero.
The applied force tends to move the block right, so static friction acts left.
For equilibrium, f required = 6 N.
f maximum = μₛN = 0.5 × 2 × 10 = 10 N.
6 N is within the 10 N limit, so f = 6 N opposite to the applied force.
The applied force becomes 8 N while the maximum static friction remains 10 N. What friction acts on the block?
You recognised the required-friction decision and applied it to a new question.
Physics · Rotational Motion · about 9 min