Question

A force of 20 N making angle 30º with vertical acts on a block kept on smooth horizontal surface. Due to action of this force,  the block is pulled along the surface by 10 m. What is the work done by this force ?​

Answer 1

Answer:

• The work done (W) by the force is 173.2 Joules

Given:

1. Force applied (F) = 20 N
2. Displacement (d) = 10 m
3. Angle between F & d (θ) = 30°

Explanation:

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Work:

It is the product of Force , displacement and the cosine angle between Force and displacement. In other words It is the scalar product of Force and displacement.

Formula:

• W = F.d cosθ

Units:

• S.I Units = Joules.
• C.G.S Units = ergs

Dimensional formula:

• M L² T⁻²

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From the formula,

⇒ W = F.d cosθ

Here,

• F Denotes Force.
• d Denotes Distance.
• θ Denotes angle.

Substituting the values,

⇒ W = 20 × 10 × cos 30°

⇒ W = 200 × cos 30°

⇒ W = 200 × √(3) / 2  ∵[cos 30° = √(3) / 2]

⇒ W = 100 × √3

⇒ W = 100 × 1.732   ∵[√3 = 1.732]

⇒ W = 173.2

⇒ W = 173.2 J

∴ The work done (W) by the force is 173.2 Joules.

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Answer 2

Given ,

• Force (f) = 20 N
• Displacement (s) = 10 m
• Angle b/w force (f) and displacement (s) = 30°

We know that ,

The scalar product of force and displacement is called work done

It is denoted by " W "

$\large \mathtt{ \fbox{Work \: done \: (W) = f . d \cos( \theta) }}$

The SI unit of work done is joule or N•m

Thus ,

➡W = 20 × 10 × Cos(30)

➡W = 200 × (√3/2)

➡W = 100 × √3

➡W = 100 × 1.732

➡W = 173.2 joule

Hence , the work done is 173.2 joule