A force F = 20 + 10y acts on a particle in y-direction where F is in newton and y in meter. Work done by this force to move the particle from y = 0 to y = 1 m is :

CONCEPT:

Work done by the force in the direction of the y-axis is written as;

\(W = \int\limits_{{y_i}}^{{y_f}} {Fdy} \)

Here, W is the work done, F is the force, y_f is the final position and y_i is the initial position.

CALCULATION:

By using work done by the variable force.It 

\(W = \int\limits_{{y_i}}^{{y_f}} {Fdy} \)      -----(1)

Given: y_i = 0, yf = 1 m 

 and F = 20 + 10y

Now, On putting this value in equation (1) we have;

\(W = \int\limits_0^1 {(20 + 10y)dy = \left[ {20y + \frac{{10{y^2}}}{2}} \right]} _0^1 = 25\,J\)

⇒ \(W { = \left[ {20y + \frac{{10{y^2}}}{2}} \right]} _0^1 \)

⇒ \(W = 25~~J\)

Hence, option 3) is the correct answer.