Question

A particle is acted upon by constant forces (4i +j-3k) and (3i+j-k) is displaced from the point (i+2j+3k) to (5i+4j+k).What is the total work done by the forces.

Answer pls test tomorrow.:(

Answer 1

Answer

Refer to the attached file

Answer 2

The workdone by the forces on the particle is 24J.

A particle is acted upon by constant forces (4i + j - 3k) and (3i + j - k) is displaced from the point (i + 2j + 3k) to (5i + 4j + k).

We have to find the total work done by the forces.

We know, work is said to be done if an object moves a certain distance due to a force acting on it. It is the dot product of net force acting on the object and the displacement of the object.

$W=\vec{F}.\vec{S}$

Here F is net force and the S is the displacement.

Given,

net force, F = F₁ + F₂

= (4i + j  - 3k) + (3i + j - k)

= 7i + 2j - 4k

displacement, S = final position - initial position

= (5i + 4j + k) -  (i + 2j + 3k)

= 4i + 2j - 2k

Now, workdone by forces , W = F.S

= (7i + 2j - 4k) . (4i + 2j - 2k)

= 28 + 4 - 8

= 24 J

Therefore the workdone by the forces is 24 J.

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