NEET + JEEAdvanced
A block of mass 2 kg rests on top of a larger block of mass 8 kg,…
Question
A block of mass 2 kg rests on top of a larger block of mass 8 kg, which itself rests on a frictionless floor. The coefficient of friction between the two blocks is 0.40. A horizontal force is applied to the upper block. Taking g = 10 m/s^2, find the maximum magnitude (in newtons) of this applied force for which the two blocks still move together without relative sliding.
✓ Verified answer: 10checked by our engine — not a guess
Step-by-step solution
The only horizontal force on the lower block is friction from the upper block, with maximum value f_max = mu*m*g = 0.4*2*10 = 8 N.
This gives the lower block's maximum acceleration while moving together: a_max = f_max/M = 8/8 = 1 m/s^2.
Moving together, the whole system (mass 10 kg) has this acceleration when F = (m+M)*a_max = 10*1 = 10 N.
Beyond this the upper block slips.
Final answer10
Stuck on a problem like this?
Paste any JEE or NEET question — verified working, a confidence %, and an honest “not sure” instead of a bluff.
Solve my doubt →More Neet Mechanics solutions
A projectile is launched from the foot of a fixed incline that rises…NEET + JEE · PhysicsOn a horizontal table, block A (2 kg) and block B (3 kg) rest in…NEET + JEE · PhysicsA ball of mass 0.5 kg is tied to a light string of length 2 m and…NEET + JEE · PhysicsA block starts from rest and slides down a smooth incline, descending…NEET + JEE · PhysicsA uniform solid sphere is released from rest at the top of a rough…NEET + JEE · PhysicsA particle executes simple harmonic motion of period 12 s about a…NEET + JEE · PhysicsA satellite of mass 200 kg moves in a circular orbit around the Earth…NEET + JEE · PhysicsA bullet of mass 0.02 kg moving horizontally at 400 m/s strikes and…NEET + JEE · Physics