Question

Two forces \(4\widehat i + \widehat j - 3\widehat k\) and \(3\widehat i + \widehat j - \widehat k\) are acting on a particle and hence the particle is displaced from the point \(\widehat i + 2\widehat j + 3\widehat k\) to the point \(5\widehat i + 4\widehat j + \widehat k\). The total work done by the forces is equal to:
1. 20 units
2. 30 units
3. 40 units
4. 50 units

Answer 1

Correct Answer - Option 3 : 40 units

Concept:

If O is the origin and \(\vec A \; and \;\vec B \) are any non-zero vector then.

\(\vec {AB}\) = position vector of \(\vec B\) - position vector of \(\vec A\)

\(\vec {AB} = \vec {OB} - \vec {OA} ....(1)\)

and total workdone by the forces \(W = (\vec {F_1} + \vec {F_2}). \vec {AB} ....(2)\)

Calculation:

Given \(\vec {OA} = \widehat i + 2\widehat j + 3\widehat k\) , \(\vec {OB} = 5\widehat i + 4\widehat j + \widehat k\)

then from equation (1), \(\vec {AB} = \vec {OB} - \vec {OA}\)

\(\vec {AB} = (5\widehat i + 4\widehat j + \widehat k) - (\widehat i + 2\widehat j + 3\widehat k)\)

\(\vec {AB} = 4 \hat i+ 2 \hat j - 2 \hat k\)

also given, \(\vec {F_1} =4\widehat i + \widehat j - 3\widehat k\) and \(\vec F_2 =3\widehat i + \widehat j - \widehat k\)

\(\vec F_1 + \vec F_2 = 7 \hat i + 2 \hat j - 4 \hat k\)

by using equation II, 

Total workdone \(W = (\vec {F_1} + \vec {F_2}). \vec {AB} \)

⇒ \(W = (7 \hat i + 2 \hat j - 4 \hat k) . (4 \hat i+ 2 \hat j - 2 \hat k)\)

⇒ W = 28 + 8 + 8 = 40 units