A constant force moves an object along the line segment from to . Find the work done if the distance is measured in meters and the force in newtons.

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A constant force F = 6i+8j-6k moves an object along a straight line from point (6, 0, -10) to point (-6, 7, 2).

Find the work done if the distance is measured in meters and the magnitude of the force is measured in newtons.

the work done is -88 J

Explanation:

Given the data in the question;

we know that;

Work done = F × S

where constant force F = ( 6i + 8j - 6k )

S = ( -6i + 7j + 2k ) - ( 6i + 0j - 10k )

S = ( (-6i - 6i) + (7j - 0j) + ( 2k - ( -10k) ) )

S = ( -12I + 7j + 12k )

so

Work force = ( 6i + 8j - 6k ) × ( -12I + 7j + 12k )

Work force = ( 6 × -12 ) + ( 8 × 7 ) + ( -6 × 12 )

Work force = -72 + 56 - 72

Work force = -88 J

Therefore, the work done is -88 J

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weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards