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On the horizontal surface of a truck a block of mass 1 kg is placed (μ=0.6) and the truck is moving with acceleration 5 m/s². Then the frictional force on the block will be

Asked in AIPMT 2001 · Friction on belts and accelerating surfaces

Answer: (1) 5 N

Step-by-step solution

Given: m=1 kg on a truck accelerating at 5 m/s²; μ=0.6.

Idea: static friction is a "supply on demand" force — it takes whatever value keeps the block moving with the truck, up to its own limit.

Friction needed to carry the block along: f=ma=1×5=5 N.

Greatest friction available: fₘₐₓ=μ mg=0.6×1×9.8=5.88 N.

Since 5<5.88, the block does not slip.

So the friction acting is exactly what is required: 5 N.

The 5.88 N is the ceiling, not the reading.

Why the other options are wrong

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