Numerical Problem No. 2.8 (Vectors and Equilibrium) Physics HSSC - I (Class -11) (Punjab Curriculum and Text Board, Lahore) With Verified Answers

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

Given:

\to F

$\vec {F}$ = 3

ˆ i

$\hat {i}$ + 2

ˆ j

$\hat {j}$

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

\to F

$\vec {F}$ and displacement

\to d

$\vec {d}$

W =

\to F

$\vec {F}$ .

\to d

$\vec {d}$ ------- 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.

\to d

$\vec {d}$ = (x₂ - x₁)

ˆ i

$\hat {i}$ + (y₂ - y₁)

ˆ j

$\hat {j}$

by putting values

\to d

$\vec {d}$ = (6 - 2)

ˆ i

$\hat {i}$ + (4 - (-1))

ˆ j

$\hat {j}$

\to d

$\vec {d}$ = 4

ˆ i

$\hat {i}$ + 5

ˆ j

$\hat {j}$

Now Eqn (1)

W =

\to F

$\vec {F}$ .

\to d

$\vec {d}$

By putting values

W = (3

ˆ i

$\hat {i}$ + 2

ˆ j

$\hat {j}$ )( 4

ˆ i

$\hat {i}$ + 5

ˆ j

$\hat {j}$ )

W = (3 ｘ 4)

ˆ i

$hat {i}$ .

ˆ i

$hat {i}$ + (3 ｘ 5)

ˆ i

$hat {i}$ .

ˆ j

$hat {j}$ + (2 ｘ 4)

ˆ j

$hat {j}$ .

$hat {i}$ + (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
⇑

List of 11th Physics Numerical Problems of Chapter - 2

⇑

Numerical Problem 2.7 ⇚ 11th Physics Main Content List ⇛ Numerical Problem 2.9

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