Question

A body of mass 40kg is placed on a horizontal surface whose coefficient of friction is 0.5 . The frictional force acting on it is ( take g =10m)

Answer 1

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Since $\large P$ is the force which is just enough to start the motion of the body,

$\large P = μ_{s}N = μ_{s}mg$

When the body moves under this force, the net force is given by

$\large F= ma = P–f$

$\large ⟹ ma = μ_{s}mg – μ_{k}N$

$\large = m(μ_{s} – μ_{k}) g$

$\large ⟹ a = (μ_{s} – μ_{k}) g$

$\large = (0.5–0.4)(10) m/s^{2}$

$\large ⟹ a = 1 m/s^{2}$

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Answer 2

Answer:

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answer

Since \large PP is the force which is just enough to start the motion of the body,

\large P = μ_{s}N = μ_{s}mgP=μ

s

N=μ

s

mg

When the body moves under this force, the net force is given by

\large F= ma = P–fF=ma=P–f

\large ⟹ ma = μ_{s}mg – μ_{k}N⟹ma=μ

s

mg–μ

k

N

\large = m(μ_{s} – μ_{k}) g=m(μ

s

–μ

k

)g

\large ⟹ a = (μ_{s} – μ_{k}) g⟹a=(μ

s

–μ

k

)g

\large = (0.5–0.4)(10) m/s^{2}=(0.5–0.4)(10)m/s

2

\large ⟹ a = 1 m/s^{2}⟹a=1m/s

2

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