Question

Calculate the work done by a force of 3 kgf in lifting a load of 1 kgf to a height of 5m.
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Answer 1

Explanation:

Given,

Force = 3 kgf

Load = 1 kgf

Height = 5 m

1 kgf = 9.8 N

1 kgf = 1 kg

Formula for work done is,

Work = Force × Displacement

W = F × D × cos(x)

x is angle of Displacement = 90 degree.

Displacement = 5 m.

Cos(90) = 0

Net force = 0 as change in KE = 0

Thus,

W = 3 × 5 × cos(90)

W = 0 J

Work done by net force is 0.

Work done against gravity = - (mg)h = -(3×9.8)×5 = -147 J

Work done in lifting object by hand to (F=mg) = (mg)h = (3× 9.8) × 5 = 147 J

Answer 2

Answer:

$\mathfrak{\underline{\underline{\red{Given: }}}}$

Force = 3 kgf

Load = 1 kgf

Height = 5 m

1 kgf = 9.8 N

1 kgf = 1 kg

Formula for work done is,

Work = Force × Displacement

W = F × D × cos(x)

x is angle of Displacement = 90 degree.

Displacement = 5 m.

Cos(90) = 0

Net force = 0 as change in KE = 0

Thus,

W = 3 × 5 × cos(90)

W = 0 J

Work done by net force is 0.

Work done against gravity = - (mg)h = -(3×9.8)×5 = -147 J

Work done in lifting object by hand to (F=mg) = (mg)h = (3× 9.8) × 5 = 147 J

Explanation:

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