Question

1) A force of '6" "m.' acting on a body at an angle of '30^(@)' withhorizontal direction displaces it horizontally through a distance of '5" "N' Calculate the work done.​

Answer 1

Given

• Force = 5N
• Displacement = 6m
• Angle between the force and the displacement is 30°

To Find

• Work Done

Solution

☯ Work = Fs cosθ

• Here as the value of θ is 30°, then the value of cosθ = cos 30° which is √3/2

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✭ According to the Question :

→ Work = Fs cosθ

• F = Force = 5 N
• s = Displacement = 6 m
• θ = Angle between the force & the Displacement = 30°

→ Work = 5 × 6 × cos 30°

→ Work = 5 × 6 × √3/2

→ Work = 30 × √3/2

→ Work = 15 × √3

→ Work = 15√3 J

or

→ Work = 15 × 1.73

→ Work = 25.95 J

∴ The work done by the force is 25.95 J

Answer 2

Question

A force of 5 N acting on a body at an Angle of 30° with horizontal direction displaces it horizontally through a distance of 6 m Calculate the work done.​

$\rule{250}{2}$

We know : Work Done = Fs × cosθ

Where

• F = Force
• s = Displacement
• θ = Angle Formed

Work Done = Fs × cosθ

⇒ Work Done = F × s × cosθ

⇒ Work Done = 5 × 6 × cosθ

⇒ Work Done = 30 × cosθ

Note : We will Have θ = 30° as it is given Angle Formed is equal to 30°

⇒ Work Done = 30 × cos 30°

⇒ Work Done = 30 × $\rm{\dfrac{\sqrt{3}}{2}}$

⇒ Work Done = 15 × √3

⇒ Work Done = 15 × 1.732

⇒ Work Done = 25.98 J