Question

Answer

Given -

F_kFk​ = 60N

m = 20 kg

g = 10 m/s²

where -

\longrightarrow⟶ F_kFk​ is the kinetic friction.

\longrightarrow⟶ m is mass of body.

\longrightarrow⟶ g is acceleration due to gravity.

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To find -

Coefficient of kinetic friction \longrightarrow⟶ \muμ

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Formula used -

\boxed{\bf \red {F_k = \mu N}}Fk​=μN​

\boxed{\bf \pink {N = mg }}N=mg​

where

\longrightarrow⟶ F_kFk​ is the kinetic friction.

\longrightarrow⟶ m is mass of body.

\longrightarrow⟶ g is acceleration due to gravity.

\longrightarrow⟶ N is normal force.

\longrightarrow⟶ \muμ is Coefficient of kinetic friction

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Solution -

m = 20 kg

g = 10 m/s²

\bf \red{N = mg}N=mg

\implies⟹ \bf N = 20 \times 10N=20×10

\implies⟹ \bf N = 200N=200

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F_kFk​ = 60N

N = 200

\bf \red{F_k = N \mu}Fk​=Nμ

\implies⟹ \bf 60 = 200 \mu60=200μ

\implies⟹ \bf \mu = \frac{200}{60} = 0.3μ=60200​=0.3

Coefficient of static friction = 0.3​

Answer 1

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