Answer
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Hint:In physics, when a body is applied with some definite magnitude of force it causes the body to displace and it covers some distance, in order to cover this distance body has some form of energy in it this simple amount of energy needed to cause a displacement of body is called work done by the body.
Complete step by step answer:
(a) When a non-zero magnitude of force is applied on the body and it covers a non-zero magnitude of distance with the condition that the direction of force and displacement must not be ${90^ \circ }$ , then it’s said that work is done on a body. Work is a scalar quantity.
Let us suppose the body moves through a distance of height $h$ while moving up against the gravity now the direction of force of gravity is $ - mg$ since it’s in downward direction and the distance is in upward direction so, Force and distance makes an angle of ${180^0}$ .
Work done is calculated as: $W = F.S.\cos (\theta )$
Where, $F$ is the magnitude of force equals to $ - mg$, $S$ Is the distance it covers which is equals to $h$ and $\theta $ Is the angle between force and displacement which is equals to ${180^0}$.
So, $W = - mgh\cos ({180^0})$
$\therefore W = mgh$
Hence, the work done by the body while moving up against gravity is $W = mgh$.
(b) Since, magnitude of force is given that $F = 2N$
Displacement $S = 10\,cm = 0.1\,m$
Since both are in same direction, so $\cos {0^0} = 1$
$W = F.S$
$\Rightarrow W = 2 \times 0.1$
$\therefore W = 0.2\,J$
Hence, the work done by the body is $W = 0.2\,J$.
Note:Remember that, Force is a vector quantity and displacement is also a vector quantity since, we take the dot product of both force and displacement hence, work done on the body is always a scalar quantity and SI unit of work I same as that of energy which is Joules.
Complete step by step answer:
(a) When a non-zero magnitude of force is applied on the body and it covers a non-zero magnitude of distance with the condition that the direction of force and displacement must not be ${90^ \circ }$ , then it’s said that work is done on a body. Work is a scalar quantity.
Let us suppose the body moves through a distance of height $h$ while moving up against the gravity now the direction of force of gravity is $ - mg$ since it’s in downward direction and the distance is in upward direction so, Force and distance makes an angle of ${180^0}$ .
Work done is calculated as: $W = F.S.\cos (\theta )$
Where, $F$ is the magnitude of force equals to $ - mg$, $S$ Is the distance it covers which is equals to $h$ and $\theta $ Is the angle between force and displacement which is equals to ${180^0}$.
So, $W = - mgh\cos ({180^0})$
$\therefore W = mgh$
Hence, the work done by the body while moving up against gravity is $W = mgh$.
(b) Since, magnitude of force is given that $F = 2N$
Displacement $S = 10\,cm = 0.1\,m$
Since both are in same direction, so $\cos {0^0} = 1$
$W = F.S$
$\Rightarrow W = 2 \times 0.1$
$\therefore W = 0.2\,J$
Hence, the work done by the body is $W = 0.2\,J$.
Note:Remember that, Force is a vector quantity and displacement is also a vector quantity since, we take the dot product of both force and displacement hence, work done on the body is always a scalar quantity and SI unit of work I same as that of energy which is Joules.
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