Question

A body of 10 kg is acted by a force of 129.4 N if $g = 9.8 \mathrm {~m} / \mathrm { sec } ^ { 2 }$ . The acceleration of the block is $10 \mathrm {~m} / \mathrm { s } ^ { 2 }$ . What is the coefficient of kinetic friction

• A. 0.03
• B. 0.01
• C. 0.30
• D. 0.25

Answer 1

Answer: (C)

0.30

Answer 2

Net force on the body = Applied force – Friction

$m a = F - \mu _ { k } m g$ ⇒ $\mu _ { k } = \frac { F - m a } { m g } = \frac { 129.4 - 10 \times 10 } { 10 \times 9.8 } = 0.3$