NEET2019Physics-Work, Energy and Power
NEET 2019 Physics Work Done by Variable Force MCQ Question
Type: MCQ-numerical-Medium-Class 11
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:
A
25 J
B
20 J
C
30 J
D
5 J
Correct Answer
Option A
Detailed Explanation
Work done is the integral of force over distance. Integrating F = 20 + 10y from 0 to 1 gives 25 J.
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