A $40 \,kg$ slab rests on a frictionless floor as shown in the figure. A $10 \,kg$ block rests on the top of the slab. The static coefficient of friction between the block and slab is $0.60$ while the kinetic friction is $0.40$. The $10\, kg$ block is acted upon by a horizontal force $100 \,N$. If $g = 9.8\,m/{s^2}$, the resulting acceleration of the slab will be ........ $m/s^2$
A $40 \,kg$ slab rests on a frictionless floor as shown in the figure. A $10 \,kg$ block rests on the top of the slab. The static coefficient of friction between the block and slab is $0.60$ while the kinetic friction is $0.40$. The $10\, kg$ block is acted upon by a horizontal force $100 \,N$. If $g = 9.8\,m/{s^2}$, the resulting acceleration of the slab will be ........ $m/s^2$
$0.98$
$1.47$
$1.52$
$6.1$
A $40 \,kg$ slab rests on a frictionless floor as shown in the figure. A $10 \,kg$ block rests on the top of the slab. The static coefficient of friction between the block and slab is $0.60$ while the kinetic friction is $0.40$. The $10\, kg$ block is acted upon by a horizontal force $100 \,N$. If $g = 9.8\,m/{s^2}$, the resulting acceleration of the slab will be ........ $m/s^2$
Limiting friction between block and slab$ = {\mu _s}{m_A}g$ $ = 0.6 \times 10 \times 9.8 = 58.8N$
But applied force on block $A$ is $100\,N$. So the block will slip over a slab.
Now kinetic friction works between block and slab ${F_k} = {\mu _k}{m_A}g$ $ = 0.4 \times 10 \times 9.8 = 39.2\;N$
This kinetic friction helps to move the slab
$\therefore $ Acceleration of slab$ = \frac{{39.2}}{{{m_B}}} = \frac{{39.2}}{{40}} = 0.98\;m/{s^2}$
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