Potential energy of a particle along x− axis is given by U=[−20+(x−2)2]J. Force on the particle at x=0 is
A
6N
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B
4N
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C
8N
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D
10N
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Solution
The correct option is B4N We know, relation between Potential energy (U) and Force (F) is - F=−dUdx........(1) Given, U=−20+(x−2)2 ∴dUdx=0+2(x−2)........(2) Putting (2) in (1), F=−2(x−2) at x=0 F=−2(0−2)=4N Force on the particle at x=0 is 4N