Question

A block of mass 2 kg is placed on the floor. The coefficient of static friction is 0.4. If a force of 2.8 N is applied on the block parallel to the floor, the force of friction between the block and the floor (taking g = 10 ms^–2) is –

• A. 2.8 N
• B. 8 N
• C. 2 N
• D. Zero

Answer 1

Answer: (A)

2.8 N

Answer 2

$\mu = 0 . 4 \longdiv { 2 \mathrm { kg } } \longrightarrow 2 . 8 \mathrm { N }$

Q 2.8 N is less than limiting force

\ static friction f_S = 2.8 N

f_L = (0.4) (2g) = 8 N