Practicing Success

Target Exam

CUET

Subject

Physics

Chapter

Work Power Energy

Question:

A force F = 20 + 10 y acts on a particle in ydirection 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

Options:

20 J

30 J

5 J

25 J

Correct Answer:

25 J

Explanation:

Work done : W = \(\int_{y_1}^{y_2}\)F.dy 

W = \(\int_{0}^{1}d\)(20 + 10y) dy = 25 J