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Q. A $2.5\, kg$ block is initially at rest on a horizontal surface. A $6.0\, N$ horizontal force and a vertical force $\vec{ P }$ are applied to the block as shown in figure. The coefficient of friction for the block and surface is $0.4$. The magnitude of friction force when $P=9 \,N$ $\left( g =10\, m / s ^{2}\right)$Physics Question Image

Laws of Motion

Solution:

$f_{\text {lim }}=\mu N=0.4(25-9)=6.4 \,N$
External force is $6 \,N$ and so block will not move .
So frictional force $=6.0 \,N$.