A block of mass $m = 2$ kg is pulled by a force $F = 40$ N upwards through a height $h = 2 m$ . Find the work done on the block by the applied force F and its weight mg. $(g = 10 m / s^{2})$

W_{F} = 80 J; W_{m g} = - 40 J

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W_{F} = 80 J; W_{m g} = 40 J

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W_{F} = - 80 J; W_{m g} = - 40 J

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W_{F} = - 80 J; W_{m g} = 40 J

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Solution

The correct option is A $W_{F} = 80 J; W_{m g} = - 40 J$
Weight $m g = (2) (10) = 20$ N

Work done by the applied force $W_{F} = F h cos 0^{\circ}$

As the angle between force and displacement is $0^{\circ}$

or $W_{F} = (40) (2) (1) = 80 J$

Similarly, work done by its weight

$W_{m g} = (m g) (h) cos 180^{\circ}$

or $W_{m g} = (20) (2) (- 1) = - 40 J$

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