54. A block of wood is placed on a rough horizontal surface. If the coefficient of static friction between the block and the surface is 0.4, what minimum force must be applied to start moving the block if its weight is 50 N?
Key Concept: Friction, Surface Irregularities, Real-World Application
b) 20 N
[Solution Description]
To move the block, the applied force must overcome the static friction. The formula for maximum static friction is $f_s = \mu_s \times N$, where $\mu_s$ is the coefficient of static friction and $N$ is the normal force. Here, the normal force equals the weight of the block, i.e., $N = 50\, \text{N}$. Substituting $\mu_s = 0.4$, we get $f_s = 0.4 \times 50 = 20\, \text{N}$. Thus, the minimum force required to start moving the block is 20 N.
Your Answer is correct.
b) 20 N
[Solution Description]
To move the block, the applied force must overcome the static friction. The formula for maximum static friction is $f_s = \mu_s \times N$, where $\mu_s$ is the coefficient of static friction and $N$ is the normal force. Here, the normal force equals the weight of the block, i.e., $N = 50\, \text{N}$. Substituting $\mu_s = 0.4$, we get $f_s = 0.4 \times 50 = 20\, \text{N}$. Thus, the minimum force required to start moving the block is 20 N.