Question

A block of mass 5 kg is on a rough horizontal surface and is at rest. Now a force of 24 N is imparted to it with negligible impulse. If the coefficient of kinetic friction is 0.4 and $g = 9.8 \mathrm {~m} / \mathrm { s } ^ { 2 }$ , then the acceleration of the block is

• A. $0.26 m / s ^ { 2 }$
• B. $0.39 \mathrm {~m} / \mathrm { s } ^ { 2 }$
• C. $0.69 \mathrm {~m} / \mathrm { s } ^ { 2 }$
• D. $0.88 \mathrm {~m} / \mathrm { s } ^ { 2 }$

Answer 1

Answer: (D)

$0.88 \mathrm {~m} / \mathrm { s } ^ { 2 }$

Answer 2

Net force = Applied force – Friction force

$m a = 24 - \mu m g$ $= 24 - 0.4 \times 5 \times 9.8$ $= 24 - 19.6$

⇒ $a = \frac { 4.4 } { 5 } = 0.88 \mathrm {~m} / \mathrm { s } ^ { 2 }$