Question

a particle Move from position 3 i^ +2j^ - 6k^ to 14i^+13j^+9k^ due to force of f =4i^+j^+3k^. if displacement in meter the work done will be​

Answer 1

A particle moved from position A ( 3i + 2j - 6k ) to position B ( 14i + 13j + 9k ) due to force F = 4i + j + 3k .

• Displacement is in meter.

We have to find the work done.

Now, We know the difference in final position and initial position will be the required displacement, but positions are in vector form. Hence,

⇒ Displacement Vector = B - A

⇒ Displacement = (14i + 13j + 9k) - (3i + 2j - 6k)

⇒ Displacement = (14 - 3)i + (13 - 2)j + (9 - (-6))k

⇒ Displacement = 11i + 11j + 15k

Now, Work done is given by scalar product of Displacement vector and Force vector. So,

⇒ Work Done = F . D

⇒ Work Done = (4i + j + 3k) . (11i + 11j + 15k)

⇒ Work Done = (4)(11) + (1)(11) + (3)(15)

⇒ Work Done = 44 + 11 + 45

⇒ Work Done = 55 + 45

⇒ Work Done = 100 J

Hence, The work done is 100 J.

Extra Information:

• Subtraction of vectors in i,j,k form,

Let any two vectors be A and B, where

A = (x₁)i + (y₁)j + (z₁)k , and

B = (x₂)i + (y₂)j + (z₂)k

Then, A - B is given by,

⇒ A - B = (x₁ - x₂) i + (y₁ - y₂) j + (z₁ - z₂) k

• Scalar Product of vectors in i,j,k form.

Let any two vectors be A and B, where

A = (x₁)i + (y₁)j + (z₁)k , and

B = (x₂)i + (y₂)j + (z₂)k

Then, A . B is given by

⇒ A . B = x₁x₂ + y₁y₂ + z₁z₂

Answer 2

Answer:

Given :-

• A particle move from position $\sf 3\hat{i} + 2\hat{j} - 6\hat{k}$ to $\sf 14\hat{i} + 13\hat{j} + 9\hat{k}$ due to force of f = $\sf 4\hat{i} + \hat{j} + 3\hat{k}$

To Find :-

• What is the work done.

Formula Used :-

$\longmapsto \sf\boxed{\bold{\pink{Work\: Done =\: \overrightarrow{F} . \overrightarrow{S}}}}$

Solution :-

First, we have to find the displacement :-

Given :

• B = $\sf 14\hat{i} + 13\hat{j} + 9\hat{k}$
• A = $\sf 3\hat{i} + 2\hat{j} - 6\hat{k}$

Then,

$\implies \sf Displacement =\: (14\hat{i} + 13\hat{j} + 9\hat{k}) -\: (3\hat{i} + 2\hat{j} - 6\hat{k})$

$\implies \sf Displacement =\: (14\hat{i} - 3\hat{i}) + (13\hat{j} - 2\hat{j}) + (9\hat{k} - (- 6\hat{k}))$

$\implies \sf Displacement =\: (11\hat{i}) + (11\hat{j}) + (9\hat{k} + 6\hat{k})$

$\implies \sf\bold{\green{Displacement =\: 11\hat{i} + 11\hat{j} + 15\hat{k}}}$

Now, we have to find the work done :-

Given :

• Force = $\sf 4\hat{i} + \hat{j} + 3\hat{k}$
• Displacement = $\sf 11\hat{i} + 11\hat{j} + 15\hat{k}$

According to the question by using the formula we get :

$\mapsto \sf Work\: Done =\: (4\hat{i} + \hat{j} + 3\hat{k}) . (11\hat{i} + 11\hat{j} + 15\hat{k})$

$\mapsto \sf Work\: Done =\: (4 \times 11) + (1 \times 11) + (3 \times 15)$

$\mapsto \sf Work\: Done =\: (44) + (11) + (45)$

$\mapsto \sf Work\: Done =\: 44 + 11 + 45$

$\mapsto \sf\bold{\red{Work\: Done =\: 100\: J}}$

$\therefore$ The work done is 100 J .