question_answer

A 40 kg slab rests on a frictionless floor. A 10 kg block rests on top of the slab. The static coefficient of friction between the block and the slab is 0.60 while the kinetic coefficient is 0.40. The 10 kg block is acted upon by horizontal force 100N. If\[\text{g }=\text{ 9}.\text{8 m}/{{\text{s}}^{\text{2}}}\], then resulting acceleration of the slab will be

A)  \[0.\text{98 m}/{{\text{s}}^{\text{2}}}\]           

B)         \[\text{1}.\text{47 m}/{{\text{s}}^{\text{2}}}\]

C) \[\text{1}.\text{52 m}/{{\text{s}}^{\text{2}}}\]         

D)         \[\text{6}.\text{1 m}/{{\text{s}}^{\text{2}}}\]

Correct Answer: A

Solution :

            \[F=\mu mg\]             \[=0.4\times 10\times 9.4=39.2\,\,N\] Resulting acceleration of the slab,             \[a=\frac{F}{m}=\frac{39\cdot 2}{40}=\mathbf{0}\mathbf{.98}\,\,\mathbf{m/}{{\mathbf{s}}^{\mathbf{2}}}\]