Question

2. When a force of 30 N acts on a body, the body
is displaced by 10 m in the direction inclined
at 60° to the force. Find the work done by the
force.
(Ans. 150 J)​

Answer 1

Given :

• Force (F) = 30 N
• Displacement (s) = 10 m
• Angle inclined (θ) = 60°

To FinD :

• Work Done by Force

SolutioN :

As we know that work done is given by :

$\longrightarrow \sf{W \: = \: F.s} \\ \\ \longrightarrow \sf{W \: = \: Fs \cos \theta}$

Substituting Given Values :

$\longrightarrow \sf{W \: = \: 30 \: \times \: 10 \: \cos 60^{\circ}} \\ \\ \longrightarrow \sf{W \: = \: 300 \: \cos 60^{\circ}} \\ \\ \longrightarrow \sf{W \: = \: \cancel{300} \: \times \: \dfrac{1}{\cancel{2}}} \\ \\ \longrightarrow \sf{W \: = \: 150} \\ \\ \underline{\sf{\therefore\ Work\ done\ by\ force\ is\ 150\ Joules}}$

___________________________

Additional information :

• When an object displaced in the direction of force applied, then work is said to be done.

• SI unit of Work is Joules.

• It is a scalar quantity.

Answer 2

$\underline{ \underline \orange{ \large\mathfrak{given : }}}$

• Force = 30 N
• Displacement = 10 m
• angle ∅ = 60°

$\underline{ \underline{ \large \mathfrak \blue{to \: find : }}}$

$\rightarrow$Work done in the process

$\underline {\underline{ \mathfrak{\red{ \large{solution : }}}}}$

Work done is the dot product of force and displacement .

$\boxed{ \rm \: \purple{work \: done = {force}\: . \: displacement}}$

In dot product , when dot is removed , cos ∅ comes into play .

$\rightarrow$ W = F s cos∅

where ,

• w = work done
• F = force applied
• s = displacement done
• ∅ = angle

$\implies$ W = 30 × 10 × cos 60°

$\implies \: \rm \: w = 300 \times \frac{1}{2}$

$\boxed {\large{ \rm\implies \: \pink{work \: done}= \green {150} \: \orange{joule }}}$

• Work done is a scalar quantity .

• Force and displacement are vector quantities .

• Dot product of two vectors gives a scalar quantity.