Physics

A body, when acted upon by a force of 10 kgf, gets displaced by 0.5 m. Calculate the work done by the force, when the displacement is
(i) in the direction of force,
(ii) at an angle of 60 degree with the force, and
(iii) normal to the force where g = 10Nkg^-1

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Given,

Force = 10 kgf and g = 10N per kg

So,

Force = 10 x 10 = 100 N

Displacement = S = 0.5 m

As we know,

Work done = force × displacement in the direction of force

(i) When displacement is in the direction of force

$W = F × S \\[0.5em] W = 100 \times 0.5 \\[0.5em] W = 50 J \\[0.5em]$

Therefore, work done is 50J.

(ii) When displacement is at an angle of 60° with the force

$W = F × S \space cos \space \theta \\[0.5em] W = 100 \times 0.5 \space Cos \space 60 \\[0.5em] W = 100 \times 0.5 \times 0.5 \\[0.5em] W = 25 J$

Therefore, work done is 25J.

(iii) When displacement is normal to the force

$W = F × S \space cos \space \theta \\[0.5em] W = 100 \times 0.5 \space Cos \space 90 \degree\\[0.5em] W = 100 \times 0.5 \times 0 \\[0.5em] W = 0$

Therefore, work done is 0.

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