Laws of MotionMedium
Question
A 40 kg slab rests on a frictionless floor. A 10kg block rests on of the slap.The static coefficient of friction between the block and slap is 0.06 while the kinetic coefficient is 0.40. The 10kg block is acted upon by a horizontal force of 100n. If g = 9.8 m/s2, resulting acceleration of the slab will be
Options
A.0.98 m/s2
B.1.47 m/s2
C.1.52 m/s2
D.6.1 m/s2
Solution
Limiting friction
Fs = μmg = 0.6 × 10 × 9.8 = 58.8N
100 N > 58.8 N
i.e slab will accelerate with different acceleration.
f = 40 a
0.4 × 10 × 9.8 = 40a ⇒ a = 0.98 m/s2
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