The force on a particle as a function of displacement x (in x-direction) is given by $ F = 10 + 0.5x $ . The work done corresponding to displacement of particle from $ x = 0 $ to $ x = 2 $ unit is
(A) $ 25J $
(B) $ 29J $
(C) $ 21J $
(D) $ 18J $

Hint
In the question the equation of force and the displacement are given. We know that work can be measured by the product of the force applied and displacement of the body in the direction of force. By applying these we can get the answer.
 $ \Rightarrow dW = Fdx $

Complete step by step answer
When a force $ \overrightarrow F $ is applied on a body which results in the displacement $ \Rightarrow \overrightarrow x $ , then the work done is the dot product of force and displacement $ W = \overrightarrow {F.} \overrightarrow x $
Here, the applied force on the particle is given as $ F = 10 + 0.5x $
Let the small displacement be dx
Then work done by the force on small displacement is $ dW = Fdx $
Integrating for $ x = 0 $ to $ x = 2 $ , we get the work done by substituting the value of F and limits of x as $ W = \int {F.dx = \int\limits_0^2 {(10 + 0.5x)dx} } = 10(2) + 0.5({2^2}/2) = 21J $
Hence, the work done is $ 21J $
The correct option is (C); $ 21J $

Additional Information
There are few points to be kept in mind -work done is independent of time and it is defined for a displacement. Work done is also independent of type of motion. The conditions that need to be fulfilled for work done are – there must be an applied force present and second is the force must produce displacement in any direction except perpendicular.

Note
If displacement is not in the direction but perpendicular to the force then no work is done. This can be seen in the case of a simple pendulum where tension in string is perpendicular to the displacement, hence the work done by tension in the string is zero. Also, it has to be kept in mind that work is a scalar quantity in spite of being a product of two vector quantities force and displacement.