Question

A force 10N acts on the body and body is displaced by 10m . The angle between force and displacement are given below .
Calculate the work done .
Angle = 30 , 45 , 60 , 90 .
Work done = ? , ? , ? , ? .

I want the answer with explanation .

Whoever gives the correct answer gets free points , a follow and I will mark them brainliest . ​

Answer 1

Explanation:

I think now u can find others when theta is 45°....

cos30°=√3/2

cos45°=1/√2

cos60°=1/2

cos90°=0

Answer 2

Given:

• Force = 10 N
• Displacement = 10 metres
• Angles are 30° , 45°, 60° , 90°.

To find:

• Work done in each cases ?

Calculation:

Work done by a force is equal to the scalar product between the force vector and the displacement vector.

General equation for work is :

$\boxed{W = f \times d \times \cos( \theta) }$

• 'f' is force, 'd' is displacement and $\theta$ is the angle.

When angle is 30°:

$W = 10 \times 10\times \cos( {30}^{ \circ} )$

$\implies W = 10 \times 10\times \dfrac{ \sqrt{3} }{2}$

$\implies W = 50 \sqrt{3} \: joule$

When angle is 45°:

$W = 10 \times 10\times \cos( {45}^{ \circ} )$

$\implies W = 10 \times 10\times \dfrac{ 1 }{ \sqrt{2} }$

$\implies W = \dfrac{ 100 }{ \sqrt{2} }$

$\implies W = 50 \sqrt{2} \: joule$

When angle is 60°:

$W = 10 \times 10\times \cos( {60}^{ \circ} )$

$\implies W = 10 \times 10\times \dfrac{ 1 }{2 }$

$\implies W = 50 \: joule$

When angle is 90°:

$W = 10 \times 10\times \cos( {90}^{ \circ} )$

$\implies W = 10 \times 10\times 0$

$\implies W = 0 \: joule$

Hope It Helps