(1 point) An object moves along a straight track from the point (βˆ’5,4,3)(βˆ’5,4,3) to the point (βˆ’2,13,βˆ’6)(βˆ’2,13,βˆ’6). The only force acting on it is a constant F=βˆ’2i+2j+3kF=βˆ’2i+2j+3k newtons. Find the work done if the distance is measured in meters. Work = joules.

Respuesta :

Answer:

-15Nm

Explanation:

Work is said to be done when a force causes an object to change its position.

Workdone = Force Γ— Distance

Given the force (βˆ’2i+2j+3k)N

Since the body moves from point (βˆ’5,4,3) to the point (βˆ’2,13,βˆ’6), we will take the difference of both points to get the distance covered.

The difference will be final point (-2,13,-6) minus initial point (-5,4,3) i.e (-2,13,-6) - (-5,4,3) = (3,9,-9)

This point can be represented vectorially as (3i+9j-9k)meters

Work done = (βˆ’2i+2j+3k)Γ—(3i+9j-9k)

Note that i.i = j.j = k.k = 1, the multiplication of different components will give "zero"

Work done = (-2Γ—3)+(2Γ—9)+(3Γ—-9)

Work done = -6+18-27

Work done= -15Nm