# Question bdbbe

Aug 27, 2017

Newton's second law of motion can be written as

$F = m a$ .......(1)

Let ${m}_{p}$ be mass of penguin, ${m}_{s}$ be mass of sled and $g$ be acceleration due to gravity.

Force of static friction between penguin and sled is

${F}_{s} = {\mu}_{s} {m}_{p} g$ .......(2)

Using (1), acceleration of penguin when it does not slide from the sled be

a = µ_s g .......(3)

The sled must also accelerates at the same rate. In such a case net force on the sled is

${F}_{\text{net}} = \left({m}_{p} + {m}_{s}\right) a$

Using (3)

${F}_{\text{net}} = \left({m}_{p} + {m}_{s}\right) \times {\mu}_{s} g$
$\implies {F}_{\text{net}} = \left({m}_{p} g + {m}_{s} g\right) \times {\mu}_{s}$
$\implies {F}_{\text{net}} = \left({w}_{p} + {w}_{s}\right) {\mu}_{s}$ ......(4)

The force of kinetic friction required to be overcome is

 F_k = µ_k (m_pg + m_sg)
 =>F_k = µ_k (w_p + w_s) ........(5)

The maximum applied force so that penguin does not slide from the sled and force of kinetic friction is also overcome is

${F}_{\max} = {F}_{\text{net}} + {F}_{k}$

Using (4) and (5)

F_max= (w_p + w_s)µ_s + (w_p + w_s)µ_k
F_max = (µ_s + µ_k)(w_p + w_s)#