Random Variables

Step 1: Recall the Formula for Work Done

The work done ( $W$ ) by a variable force $F(y)$ over a displacement from $y_1$ to $y_2$ is given by the integral:

$W%20%3D%20%5Cint_%7By_1%7D%5E%7By_2%7D%20F(y)%20%5C%2C%20dy$

Step 2: Substitute the Given Force and Limits

In this case, $F(y)%20%3D%205%20%2B%203y%5E2$ , $y_1%20%3D%202%20%5C%2C%20%5Ctext%7Bm%7D$ , and $y_2%20%3D%205%20%5C%2C%20%5Ctext%7Bm%7D$ . So,

$W%20%3D%20%5Cint_%7B2%7D%5E%7B5%7D%20(5%20%2B%203y%5E2)%20%5C%2C%20dy$

Step 3: Evaluate the Integral

$W%20%3D%20%5Cleft%5B%205y%20%2B%20%5Cfrac%7B3y%5E3%7D%7B3%7D%20%5Cright%5D_2%5E5%20%3D%20%5Cleft%5B%205y%20%2B%20y%5E3%20%5Cright%5D_2%5E5$ $W%20%3D%20(5(5)%20%2B%205%5E3)%20-%20(5(2)%20%2B%202%5E3)$ $W%20%3D%20(25%20%2B%20125)%20-%20(10%20%2B%208)$ $W%20%3D%20150%20-%2018%20%3D%20132%20%5C%2C%20%5Ctext%7BJ%7D$

Conclusion: The work done by the force is $132%20%5C%2C%20%5Ctext%7BJ%7D$ .

High-Yield Trend

Chapter Questions
1 MCQs

Official Solution

Random Variables

High-Yield Trend

Chapter Questions 1 MCQs

Official Solution

Chapter Questions
1 MCQs