Answer:
0.286
Explanation:
Let g = 10m/s2, and assume the pulley is frictionless, the weight by the 10kg block would exert a force on the 35kg block on the ground
Weight of the 10 kg-block = mg = 10 * 10 = 100 N
This force is balanced by static friction force from the ground due to the normal force of the 35kg block. So the friction force would also be 100N
The weight and normal force of the 35kg block is N = Mg = 35*10 = 350 N
The coefficient of the friction force if friction force divided by normal force:
[tex]\mu = \frac{F_f}{N} = \frac{100}{350} = 0.286 [/tex]
