A gardener pushes a lawn roller through a distance of \[20{\text{ m}}\]. If he applies a force of \[20{\text{ kg - wt}}\]in a direction inclined at \[{60^ \circ }\] to the ground, the work done by him is
A. \[1960{\text{ J}}\]
B. \[{\text{196 J}}\]
C. \[{\text{1}}{\text{.96 J}}\]
D. \[{\text{196 KJ}}\]

Hint: In this question, we need to find out the work done by the gardener while pushing a lawn roller when he applies a constant force of \[20{\text{ kg - wt}}\]in a direction inclined at \[{60^ \circ }\]. As the force given is constant, we need to use the formula \[{\text{W = FScos}}\theta \]. Convert the force in \[{\text{kg - wt}}\] to force in newton and on substituting in the formula we get work done by the gardener.

Formula used:
\[{\text{W = FScos}}\theta \].............(For constant force)
Where \[F = \] Constant force applied, \[S = \] Displacement, \[\theta = \] angle between \[{\text{F}}\] and \[{\text{S}}\]

Complete step by step answer:
Let us write down the given data, force applied by the gardener to push the lawn roller,
\[F = 20{\text{ kg - wt}}\]
\[\Rightarrow {\text{1 kg - wt}} = 9.8{\text{ N}}\]
Therefore,
\[F = 20{\text{ kg - wt }} \\
\Rightarrow F = {\text{ 20}} \times 9.8{\text{ N}} \\
\Rightarrow F = {\text{196 N}}\]
Displacement of the lawn roller,
\[{\text{S = 20 m}}\]
The Angle between the direction of force applied and direction of displacement is,
\[\theta = {60^ \circ }\]

As the force applied by the gardener is constant, we can use the formula, \[{\text{W = FScos}}\theta \] to find out the work done by the gardener
Substituting \[{\text{F = 196 N}}\], \[{\text{S = 20 m}}\], and \[\theta = {60^ \circ }\]in the formula we get,
\[W = 196 \times 20 \times \cos {60^ \circ }\]
\[ \Rightarrow \cos {60^ \circ } = \dfrac{1}{2}\]
\[ \Rightarrow W = 196 \times 20 \times \dfrac{1}{2}\]
\[ \therefore W = 1960{\text{ J}}\]
Hence, work done by the gardener in pushing the lawn roller is \[1960{\text{ J}}\].

Therefore, the correct option is A.

Additional information:Work done by a force may be positive, negative or even zero also, depending on the angle \[\theta \] between the force vector \[{\text{F}}\] and displacement vector \[{\text{S}}\]. Work done by a force is zero when \[\theta {\text{ = }}{90^ \circ }\], it is positive when \[0 \leqslant \theta \leqslant {90^ \circ }\] and negative when \[{90^ \circ } \leqslant \theta \leqslant {180^ \circ }\]. Work depends on the frame of reference. With the change of frame of reference, inertial force does not change while displacement may change. So, the work done by a force will be different in different frames.

Note: The formula \[{\text{W = FScos}}\theta \] is applicable only when the force given is constant and applied in a particular direction if the force is varying in magnitude, we cannot use dot product to find out the work done instead we need to integrate the force over the range of displacement to find out the work done. In that case, work done, \[W = \int\limits_{{s_1}}^{{s_2}} {dW} = \int\limits_{{s_1}}^{{s_2}} {F \cdot ds} \]