Friction: Exam-Pattern Drill (5)

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Question

A block of mass 55 kg rests on a horizontal surface. The coefficient of static friction is μs=0.4\mu_s = 0.4 and kinetic friction is μk=0.3\mu_k = 0.3. A horizontal force is gradually increased from 00 N. Find (a) the force at which the block just begins to slip, (b) the acceleration once it is moving, if the applied force is held at 2525 N. Take g=10g = 10 m/s2^2.

Solution — Step by Step

The block stays at rest as long as applied force ≤\leq maximum static friction. Maximum static friction:

fsmax⁡=μsmg=0.4×5×10=20 Nf_s^{\max} = \mu_s mg = 0.4 \times 5 \times 10 = 20 \text{ N}

At F=20F = 20 N, the block is on the verge of slipping. Just past this, it starts moving.

Once moving, friction drops to kinetic friction:

fk=μkmg=0.3×5×10=15 Nf_k = \mu_k mg = 0.3 \times 5 \times 10 = 15 \text{ N} Fnet=F−fk=25−15=10 NF_{\text{net}} = F - f_k = 25 - 15 = 10 \text{ N} a=Fnetm=105=2 m/s2a = \frac{F_{\text{net}}}{m} = \frac{10}{5} = 2 \text{ m/s}^2

Final Answer: (a) 2020 N to just start slipping. (b) Acceleration =2= 2 m/s2^2 once moving with F=25F = 25 N.

Why This Works

Static friction is a self-adjusting force — it matches the applied force up to its maximum μsN\mu_s N, keeping the block stationary. Beyond that threshold, the block accelerates and kinetic friction (a smaller, fixed value μkN\mu_k N) takes over.

The key conceptual point: at the instant of slipping, friction drops from 2020 N to 1515 N. So even with the same applied force just above threshold, the block experiences a sudden net force and lurches forward.

Alternative Method

Energy-method check: in 11 second of motion at a=2a = 2 m/s2^2, the block reaches v=2v = 2 m/s, KE =12(5)(4)=10= \frac{1}{2}(5)(4) = 10 J. Work by applied force minus work by friction =25×1−15×1=10= 25 \times 1 - 15 \times 1 = 10 J. (Used s=12at2=1s = \frac{1}{2}at^2 = 1 m.) Consistent.

Using μs\mu_s throughout the problem is the typical slip. Once the block moves, friction is kinetic, not static. The transition from μs\mu_s to μk\mu_k is the entire conceptual hook of this problem type.

JEE Main loves a variant where the surface is inclined at angle θ\theta. The threshold condition becomes tan⁡θ=μs\tan\theta = \mu_s (angle of repose). Memorize this — it appears in multiple-choice form almost every year.

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