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:
I don’t know sorry dear