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 ( g =10 m / s 2)

Question

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 ( g =10 m / s 2) (A) 6.0 N (B) 6.4 N (C) 9.0 N (D) zero

Tardigrade Admin · Accepted Answer

f text lim =μ 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.

Q. A $2.5 k g$ block is initially at rest on a horizontal surface. A $6.0 N$ horizontal force and a vertical force $P$ 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 $P = 9 N$ $(g = 10 m / s^{2})$

6.0 N

6.4 N

9.0 N

zero

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

Q. A 2.5kg block is initially at rest on a horizontal surface. A 6.0N horizontal force and a vertical force 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=9N (g=10m/s2)