Friction: Step-by-Step Worked Examples (3)

hard 2 min read

Question

A block of mass 5 kg5\ \text{kg} rests on a horizontal surface. The coefficients of static and kinetic friction are μs=0.4\mu_s = 0.4 and μk=0.3\mu_k = 0.3. A horizontal force is gradually increased from zero. At what applied force does the block start moving, and what is its acceleration just after motion begins? Take g=10 m/s2g = 10\ \text{m/s}^2.

Solution — Step by Step

The block stays at rest until applied force exceeds fs,maxf_{s,\max}:

fs,max=μsN=μsmg=0.4×5×10=20 Nf_{s,\max} = \mu_s N = \mu_s mg = 0.4 \times 5 \times 10 = 20\ \text{N}

The block starts sliding the instant applied force FF exceeds 20 N20\ \text{N}. So the threshold is F=20 NF = 20\ \text{N} (any infinitesimal excess starts motion).

fk=μkmg=0.3×5×10=15 Nf_k = \mu_k mg = 0.3 \times 5 \times 10 = 15\ \text{N}

The instant motion starts, friction drops from 20 N20\ \text{N} (static max) to 15 N15\ \text{N} (kinetic).

Net force just after motion begins (applied force still 20 N\approx 20\ \text{N}):

Fnet=2015=5 N,a=Fnetm=55=1 m/s2F_{\text{net}} = 20 - 15 = 5\ \text{N}, \quad a = \frac{F_{\text{net}}}{m} = \frac{5}{5} = 1\ \text{m/s}^2

Final answer: Block starts moving at F=20 NF = 20\ \text{N}. Initial acceleration is a=1 m/s2a = 1\ \text{m/s}^2.

Why This Works

The "μs>μk\mu_s > \mu_k" inequality is what gives friction its surprising behaviour: a sudden jump in available net force the moment motion begins. This is why pushing a heavy box is hardest at the start.

Drawing the friction-versus-applied-force graph (linear up to μsmg\mu_s mg, then a jump down to μkmg\mu_k mg) makes this lesson permanent.

Alternative Method

Use energy methods over a small displacement. Work done by applied force minus work by kinetic friction equals KE gained. Same answer, but slower for this simple case.

JEE Advanced loves “block on block” friction problems. Start by drawing free-body diagrams for each block separately and identifying which surface’s friction acts on which block.

Common Mistake

Using μs\mu_s in Newton’s second law after motion starts. Once the block moves, only μk\mu_k matters. Mixing them gives a=0a = 0 — a “trap” answer that suggests the block doesn’t move at all.

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