NEETPhysicsWork Energy and PowerMCQ+4 / −1
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 :
- A5 J
- B25 J
- C20 J
- D30 J
View written solutionFree
Correct answer: B
Work done under the given variable force is :
W =
Here, y1 = 0, y2 = 1 m
W =
= = 25 J
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