Question

A particle moves from point P(0, 0, 0) m to Q(0, 1, 1) m under the action of a constant force The work done by the force is

Choose answer:

2 J


–2 J


3 J


–6 J

Answer 1

$\huge{\underline{\bold{\underline{Correct question:-}}}}$

#refer the attachment for question.

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

Position is given as:-

1. P(0, 0, 0)

Which is $\large{S_1 = 0 \hat{i} + 0 \hat{j} + 0 \hat{k}}$

2. Q(0, 1, 1)

Which is $\large{S_2 = 0 \hat{i} + 1 \hat{j} + 1 \hat{k}}$

$\huge{\bold{\underline{Explanation:-}}}$

It is given that the force is :-

$\large{\bold{F = \hat{i} + 2 \hat{j} - 4 \hat{k}}}$

Firstly finding the position vector

$\large{ \Delta S = S_2 - S_1}$

Now,

$\large{ \implies \Delta S = 0 \hat{i} + 1 \hat{j} + 1 \hat{k} -( 0 \hat{i} + 0 \hat{j} + 0 \hat{k})}$

Which is equal to

$\large{ \implies \Delta S = 0 \hat{i} + 1 \hat{j} + 1 \hat{k}}$

Applying Work done Formula.

$\large{\bold{W = F. \Delta S}}$

Substituting the values,

$\large{ \implies W = 1 \hat{i} + 2 \hat{j} - 4 \hat{k} . ( 0 \hat{i} + 1 \hat{j} + 1 \hat{k})}$

We need to apply here Scalar product,

$\large{ \implies W = 1 \times 0 + 2 \times 1 - 4 \times 1}$

$\large{ \implies W = 0 + 2 - 4}$

$\large{ \implies W = 2 - 4}$

$\huge{\boxed{\boxed{W = - 2 \: J}}}$

So, the work done by the force is - 2 Joules (Option - 2).