Find the work done when the point of application of the force 3`\hat {i}` + 2`\hat {j}` moves in a straight line from the point (2,-1) to the point (6,4). (Ans: 22 units)


Given:

`\vec {F}` = 3`\hat {i}` + 2`\hat {j}` 
Point 1 = (2, -1)
Here  x₁ = 2 and  y₁ = -1
Point 2 = (6,4)
Here x₂ = 6 and y₂ = 4

To Find:

Word done = W=?

Solution: 

We know that is the scaller product of Force `\vec {F}` and displacement `\vec {d}`

W = `\vec {F}` . `\vec {d}`  ------- Eqn (1)

The value of displacement d in From Eqn (1) is also unknown. So, we will first find the displacement d from the given two points (2, -1) and (6,4) by using the below formula.

`\vec {d}` = (x₂ - x₁) `\hat {i}` + (y₂ - y₁)`\hat {j}` 

by putting values

`\vec {d}` = (6 - 2) `\hat {i}` + (4 - (-1))`\hat {j}` 

`\vec {d}` = 4 `\hat {i}` + 5`\hat {j}` 

Now Eqn (1)

W = `\vec {F}` . `\vec {d}`

By putting values

W = (3`\hat {i}` + 2`\hat {j}` )( 4 `\hat {i}` + 5`\hat {j}` )

W = (3 ｘ 4) `hat {i}` . `hat {i}` + (3 ｘ 5) `hat {i}` . `hat {j}` + (2 ｘ 4) `hat {j}` . `hat {i}` + (2 ｘ 5) `hat {j}` . `hat {j}`

[∴ `hat {i}` . `hat {i}` = `hat {j}` . `hat {j}` =1 and `hat {i}` . `hat {j}` = `hat {j}` . `hat {i}` =0 ]

So by putting these values we have

W = 12 + 0 + 0 + 10

W = 22 units--------Ans 

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Numerical Problem 2.8
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List of 11th Physics Numerical Problems of Chapter - 2
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