Question

If two forces of magnitude 7 and 50 units act in the directions 31 + 2) - 6k and 92 - 123+
20k respectively on a particle moving it from the point A (1,0,-3) to the point B (3,-2,-5), then the
work done by the forces is
A) 14 units B) 27 units C) 18 units D) 24 units​

Answer 1

Option B) 27 units is the correct answer.

Given:

• Two forces of magnitude 7 and 50 units
• Directions of the forces: 31i + 2j - 6k and 92i - 123j + 20k
• Points: A(1,0,-3) and B(3,-2,-5)

To find:

• Work done by the forces on a particle moving from point A to point B.

Solution:

We can use the formula W = F . d, where W is the work done, F is the force, and d is the displacement. To find the force, we can use the given magnitudes and directions to calculate the force vectors.

The first force has magnitude 7 and direction vector (31, 2, -6), so its force vector is 7(31i + 2j - 6k).

The second force has magnitude 50 and direction vector (92, -123, 20), so its force vector is 50(92i - 123j + 20k).

To find the displacement vector, we can subtract the initial position vector from the final position vector:

(3i - 2j - 5k) - (i + 0j - 3k)

= 2i - 2j + 2k.

Now we can calculate the work done by each force:

Work done by the first force

= F1 . d = 7(31i + 2j - 6k) . (2i - 2j + 2k)

= 14 - 28 + 84 = 70 units.

Work done by the second force

= F2 . d = 50(92i - 123j + 20k) . (2i - 2j + 2k)

= 9200 - 12300 + 2000 = 2700 units.

Therefore, the total work done by the two forces is 70 + 2700 = 2770 units, which is closest to option B) 27 units.

#SPJ1

Answer 2

EXPLANATION.

Two forces of magnitude 7 and 50.

units act in the direction $3 \hat{i} + 2 \hat{j} - 6 \hat{k}$  and  $92 \hat{i} - 123 \hat{j} + 20 \hat{k}$  respectively.

Particle moving it from point A (1, 0, -3) to the point B (3, - 2, - 5).

As we know that,

$\sf \displaystyle \vec{F} = F. \hat{n}$

$\sf \displaystyle \hat{n} = indicates \ the \ unit \ vector$

$\sf \displaystyle F = Forces$

$\sf \displaystyle \vec{F}_{1} = F \hat{n}$

$\sf \displaystyle \vec{F}_{1} = 7 \bigg(\frac{3 \hat{i} + 2 \hat{j} - 6 \hat{k}}{|3 \hat{i} + 2 \hat{j} - 6 \hat{k}|} \bigg)$

$\sf \displaystyle \vec{F}_{1} = 7 \bigg(\frac{3 \hat{i} + 2 \hat{j} - 6 \hat{k}}{\sqrt{(3)^{2} + (2)^{2} + (-6)^{2} } } \bigg)$

$\sf \displaystyle \vec{F}_{1} = 7 \bigg(\frac{3 \hat{i} + 2 \hat{j} - 6 \hat{k}}{\sqrt{9 + 4 + 36} } } \bigg)$

$\sf \displaystyle \vec{F}_{1} = 7 \bigg(\frac{3 \hat{i} + 2 \hat{j} - 6 \hat{k}}{\sqrt{49 } } \bigg)$

$\sf \displaystyle \vec{F}_{1} = 7 \bigg(\frac{3 \hat{i} + 2 \hat{j} - 6 \hat{k}}{7 } } \bigg)$

$\sf \displaystyle \vec{F}_{1} = 3 \hat{i} + 2 \hat{j} - 6 \hat{k}$

$\sf \displaystyle \vec{F}_{2} = \vec{F} \hat{n}$

$\sf \displaystyle \vec{F}_{2} = 50 \bigg(\frac{92 \hat{i} - 123 \hat{j} + 20 \hat{k}}{|92 \hat{i} - 123 \hat{j} + 20 \hat{k} |} \bigg)$

$\sf \displaystyle \vec{F}_{2} = 50 \bigg(\frac{92 \hat{i} - 123 \hat{j} + 20 \hat{k}}{\sqrt{(92)^{2} + (-123)^{2} + (20)^{2} } } \bigg)$

$\sf \displaystyle \vec{F}_{2} = 50 \bigg(\frac{92 \hat{i} - 123 \hat{j} + 20 \hat{k}}{\sqrt{8464 + 15129 + 400} } } \bigg)$

$\sf \displaystyle \vec{F}_{2} = 50 \bigg(\frac{92 \hat{i} - 123 \hat{j} + 20 \hat{k}}{\sqrt{23993} } } \bigg)$

$\sf \displaystyle \vec{F}_{2} = 50 \bigg(\frac{92 \hat{i} - 123 \hat{j} + 20 \hat{k}}{154.89 } } \bigg)$

$\sf \displaystyle \vec{F}_{2} = 50 \bigg(\frac{92 \hat{i} - 123 \hat{j} + 20 \hat{k}}{155 } } \bigg)$

$\sf \displaystyle \vec{F}_{2} = 10 \bigg(\frac{92 \hat{i} - 123 \hat{j} + 20 \hat{k}}{31 } } \bigg)$

Resultant of two forces.

$\sf \displaystyle F_{Re} = F_{1} + F_{2}$

$\sf \displaystyle F_{Re} = (3 \hat{i} + 2 \hat{j} - 6 \hat{k}) + 10 \bigg(\frac{92 \hat{i} - 123 \hat{j} + 20 \hat{k}}{31} \bigg)$

$\sf \displaystyle F_{Re} = \frac{31(3 \hat{i} + 2 \hat{j} - 6 \hat{k})}{31} + 10 \bigg(\frac{92 \hat{i} - 123 \hat{j} + 20 \hat{k}}{31} \bigg)$

$\sf \displaystyle F_{Re} = \frac{(93 \hat{i} + 62 \hat{j} - 186 \hat{k})}{31} + \frac{(920 \hat{i} - 1230 \ht{j} + 200 \hat{k})}{31}$

$\sf \displaystyle F_{Re} = \frac{(93 \hat{i} + 920 \hat{i} ) + (62 \hat{j} - 1230 \hat{j} ) + (- 186 \hat{k} + 200 \hat{k} )}{31}$

$\sf \displaystyle F_{Re} = \frac{(1013 \hat{i} - 1168 \hat{j} + 14 \hat{k})}{31}$

Now, we need to find the displacement.

$\sf \displaystyle d_{1} = (\hat{i} + 0 \hat{j} - 3 \hat{k})$

$\sf \displaystyle d_{1} = (\hat{i} - 3 \hat{k})$

$\sf \displaystyle d_{2} = (3 \hat{i} - 2 \hat{j} - 5 \hat{k})$

$\sf \displaystyle d_{Re} = d_{2} - d_{1}$

$\sf \displaystyle d_{Re} = (3 \hat{i} - 2 \hat{j} - 5 \hat{k} ) - (\hat{i} - 3 \hat{k})$

$\sf \displaystyle d_{Re} = 3 \hat{i} - 2 \hat{j} - 5 \hat{k} - \hat{i} + 3 \hat{k}$

$\sf \displaystyle d_{Re} = 2 \hat{i} - 2 \hat{j} - 2 \hat{k}$

As we know that,

Work done = Force x Displacement.

Using this formula to find the work done by the forces.

$\sf \displaystyle Work \ done = \frac{(1013 \hat{i} - 1168 \hat{j} + 14 \hat{k})}{31} \times (2 \hat{i} - 2 \hat{j} - 2 \hat{k})$

$\sf \displaystyle Work \ done = \frac{(2026 + 2336 - 28)}{31}$

$\sf \displaystyle Work \ done = \frac{(4362- 28)}{31}$

$\sf \displaystyle Work \ done = \frac{(4334)}{31}$

$\sf \displaystyle Work \ done = 139. 80 \ units$

∴ Work done by the forces is 139.80 units.