Question

A 60 kg body is pushed with just enough force to start it moving across a floor and the same force continues to act afterwards. The coefficient of static friction and sliding friction are 0.5 and 0.4 respectively. The acceleration of the body is

• A. $6 m / s ^ { 2 }$
• B. $4.9 m / s ^ { 2 }$
• C. $3.92 m / s ^ { 2 }$
• D. $1 m / s ^ { 2 }$

Answer 1

Answer: (D)

$1 m / s ^ { 2 }$

Answer 2

Limiting friction

$= \mu _ { s } R = \mu _ { s } m g = 0.5 \times 60 \times 10 = 300 N$

Kinetic friction

$= \mu _ { k } R = \mu _ { k } m g = 0.4 \times 60 \times 10 = 240 N$

Force applied on the body = 300 N and if the body is moving then,

Net accelerating force=Applied force – Kinetic friction

⇒ $m a = 300 - 240 = 60$ $\therefore$ $a = \frac { 60 } { 60 } = 1 \mathrm {~m} / \mathrm { s } ^ { 2 }$