Question

17. A block of mass 10kg is placed on rough horizontal plane and a force of 8 N is applied. If u = 0.1 the
frictional force acting on it is____
N.​

Answer 1

Answer -

Given -

m = 10 kg

$\mu$ = 0.1

g = 10 m/s²

where -

$\longrightarrow$$\mu$ is the coefficient of kinetic friction.

$\longrightarrow$m is mass of body.

$\longrightarrow$g is acceleration due to gravity.

━━━━━━━━━━━━━

To find -

Frictional force $\longrightarrow$$F$

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Formula used -

$\boxed{\bf \red {F_k = \mu N}}$

$\boxed{\bf \pink {N = mg }}$

where

$\longrightarrow$$F_k$ is the kinetic friction.

$\longrightarrow$m is mass of body.

$\longrightarrow$g is acceleration due to gravity.

$\longrightarrow$N is normal force.

$\longrightarrow$$\mu$ is Coefficient of kinetic friction

━━━━━━━━━━━━━

Solution -

N = mg

N = 10 × 10

N = 100 Newton

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F = $\mu$ N

F = 0.1 × 100

F = 10N

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Thanks

Answer 2

Explanation:

Given -

m = 10 kg

�μ = 0.1

g = 10 m/s²

where -

⟶⟶ �μ is the coefficient of kinetic friction.

⟶⟶ m is mass of body.

⟶⟶ g is acceleration due to gravity.

━━━━━━━━━━━━━

To find -

Frictional force ⟶⟶ �F

━━━━━━━━━━━━━

Formula used -

��=��Fk=μN

�=��N=mg

where

⟶⟶ ��Fk is the kinetic friction.

⟶⟶ m is mass of body.

⟶⟶ g is acceleration due to gravity.

⟶⟶ N is normal force.

⟶⟶ �μ is Coefficient of kinetic friction

━━━━━━━━━━━━━

Solution -

N = mg

N = 10 × 10

N = 100 Newton

━━━━━━━━━━━━━

F = �μ N

F = 0.1 × 100

F = 10N

━━━━━━━━━━━━━

Thanks