Laws of MotionHard
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
Create a free account to view solution
View Solution FreeMore Laws of Motion Questions
Two masses of 20 kg and 10 kg are connected by a light string passing over a frictionless pulley. The coefficient of fri...Ten one rupees coins are put on top of each other on a table. Each coin has a mass ′m′ kg., then the force o...In the figure, a man of true mass M is standing on a weighing machine placed in a cabin. The cabin is joined by a string...A man weighing 100 kg carries a load of 10 kg on his head. He jumps from a tower with the load on his head. What will be...A block of mass m lying on a rough horizontal plane is acted upon by a horizontal force P and another force Q inclined a...