**Friction force **(F_{f}) is equal to the product of **coefficient of friction **(µ) and the **normal force **(F_{N}). Using the equation of friction: **F**_{f}** = µ × F**** _{N}**, the value of friction force acting on an object can be calculated.

Let’s solve some problems based on this equation, so you’ll get a clear idea.

## Friction Practice Problems

**Problem 1:** A box of mass 9 kg is pushed along the floor. If the coefficient of friction between the box and floor is 0.09, then calculate the value of friction force acting on a box.

Solution:

Given data:

Mass of a box, m = 9 kg

Coefficient of friction between the box and floor, µ = 0.09

Friction force acting on a box, F_{f} = ?

Assume: gravitational acceleration, g = 9.81 m/s^{2}

Using the equation of friction force,

F_{f} = µ × F_{N}

F_{f} = µ × mg (equation of normal force, F_{N} = mg)

F_{f} = 0.09 × 9 × 9.81

F_{f} = 7.94 N

Therefore, the friction force acting on a box is **7.94 N**.

**Problem 2:** What is the value of friction force acting on an ice slab of mass 12 kg, when it is pulled along a cement floor. The coefficient of friction between the ice slab and cement floor is 0.35.

Solution:

Given data:

Mass of an ice slab, m = 12 kg

Coefficient of friction between the ice slab and cement floor, µ = 0.35

Friction force acting on an ice slab, F_{f} = ?

Assume: gravitational acceleration, g = 9.81 m/s^{2}

Using the equation of friction force,

F_{f} = µ × F_{N}

F_{f} = µ × mg (equation of normal force, F_{N} = mg)

F_{f} = 0.35 × 12 × 9.81

F_{f} = 41.20 N

Therefore, the friction force acting on an ice slab is **41.20 N**.

**Problem 3:** A 2 kg book resting on the desk is pushed with the hand. Calculate the friction force acting on a book, if the coefficient of friction between the book and desk is 0.24.

Solution:

Given data:

Mass of a book, m = 2 kg

Friction force acting on a book, F_{f} = ?

Coefficient of friction between the book and desk, µ = 0.24

Assume: gravitational acceleration, g = 9.81 m/s^{2}

Using the equation of friction force,

F_{f} = µ × F_{N}

F_{f} = µ × mg (equation of normal force, F_{N} = mg)

F_{f} = 0.24 × 2 × 9.81

F_{f} = 4.70 N

Therefore, the friction force acting on a book is **4.70 N**.

**Problem 4:** Calculate the friction force acting on a 500 gm knife which is sharpened on a knife sharpening stone. The coefficient of friction between a knife and knife sharpening stone is 0.75.

Solution:

Given data:

Friction force acting on a knife, F_{f} = ?

Mass of a knife, m = 500 gm = 0.5 kg

Coefficient of friction between a knife and knife sharpening stone, µ = 0.75

Assume: gravitational acceleration, g = 9.81 m/s^{2}

Using the equation of friction force,

F_{f} = µ × F_{N}

F_{f} = µ × mg (equation of normal force, F_{N} = mg)

F_{f} = 0.75 × 0.5 × 9.81

F_{f} = 3.67 N

Therefore, the friction force acting on a knife is **3.67 N**.

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- Terminal Velocity Equation
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- Inertia Formula
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