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JEE Main 2014 · 9 Apr · Shift 1, Physics question 4

A block A of mass 4 kg is placed on another block B of mass 5 kg, and the block B rests on a smooth horizontal table. If the minimum force that can be applied on A so that both the blocks move together is 12 N, the maximum force that can be applied on B for the blocks to move together will be :

Choose one option
Answer and worked solution

Official answer: C

The key's reading: the 12 N is the limiting friction between the blocks, i.e. A just starts to slide on B under 12 N.

When the force is applied to B, friction is the only horizontal force on A. So A can accelerate at most at
amax⁡=124=3 m/s2.a_{\max}=\frac{12}{4}=3\text{ m/s}^2.
For both blocks to move together,
Fmax⁡=(4+5)×3=27 N.F_{\max}=(4+5)\times3=27\text{ N}.

Topics:FrictionNewton's laws
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