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]

Question

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] (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

F.B.D. of the block is shown in the diagram

Since block is at rest therefore

f r - m g = 0 \dots

(1)

F - N = 0 \dots

(2)

f r \leq μ N

In limiting case

f r = μ N = μ F \dots

(3)
Using eq. (1) and (3)

∴ μ F = m g

\Rightarrow F = \frac{0.5 \times 10}{0.2} = 25 N

Q. The coefficient of static friction between a wooden block of mass 0.5kg 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=10ms−2]

F.B.D. of the block is shown in the diagram Since block is at rest therefore fr−mg=0…(1) F−N=0…(2) fr≤μN In limiting case fr=μN=μF…(3) Using eq. (1) and (3) ∴μF=mg ⇒F=0.20.5×10​=25N

Q. The coefficient of static friction between a wooden block of mass $0.5 k g$ 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. $[g = 10 m s^{- 2}]$

F.B.D. of the block is shown in the diagram

Since block is at rest therefore
$f r - m g = 0 \dots$ (1)
$F - N = 0 \dots$ (2)
$f r \leq μ N$
In limiting case $f r = μ N = μ F \dots$ (3)
Using eq. (1) and (3)
$∴ μ F = m g$
$\Rightarrow F = \frac{0.5 \times 10}{0.2} = 25 N$