Question

A fireman of mass $60\, kg$ slides down a pole He is pressing the pole with a force of $600\, N$. The coefficient of friction between the hands and the pole is $0.5$ with what acceleration with the fireman slide down? $\left(g=10\, m / s ^{2}\right)$

• A. $1 \,m / s ^{2}$
• B. $2.5\, m / s ^{2}$
• C. $10 \,m / s ^{2}$
• D. $5 \,m / s ^{2}$

Answer 1

Answer: (D)

Net downward acceleration

$=\frac{\text { Weight - Frictional force }}{\text { Mass }} $
$=\frac{m g-\mu R}{m} $
$=\frac{60 \times 10-0.5 \times 600}{60} $
$=\frac{300}{60}=5\, m / s ^{2}$