Question

A 40 kg slab rests on a frictionless floor. A 10 kg block rests on the top of the slab as shown in fig. 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 a horizontal force of 100 N. If g = 9.8 m $s^{-2}$, the resulting acceleration of the slab will be

• A. $0.784m{s}^{-2}$
• B. $1.52m{s}^{-2}$
• C. $1.47m{s}^{-2}$
• D. $0.98m{s}^{-2}$

Answer 1

The correct option is A $0.784m{s}^{-2}$

${\mu }_s=0.6{\left(f_s\right)}_{max}=0.6\times 10\times 18=58.8N{\mu }_k=0.4f_k=0.4\times 10\times 9.8=39.2Nso,100N>58.8N$

Hence $10$kg will move and surface will create $f_k$ force on $10$kg in right direction and hence $f_k$ force will be introduced on $40$kg slab in left direction.

Net force on slab =$f_k=39.2N$

$a_{slab}=\frac{39.2}{10+40}=0.784m{s}^{-2}$