The coefficient of static friction between a wooden block of mass 0.5 kg and a vertical rough wall is 0.2. The magnitude of horizontal force that should be applied on the block to keep it adhere to the wall will be ______ N. [g = 10 ms^{-2}]

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the correct answer is 25 N

Given:: \(\mu_s\) = 0.2

Solution::

m=0.5 kg

g=10 m/s^{2}

We know that \(f_s= \mu N\)

To keep the block adhere to the wall

here N = F ...(2)

f_{s} = mg ....(3)

from equation (1), (2), and (3), we get

\(= mg=\mu F\)

\(\therefore F={mg\over\mu}\)

\(F={0.5×10\over0.2}\)

F=25N

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