Question

In an experiment, one of the forces exerted on a proton is F⃗ =−αx2i^, where α=12N/m2. What is the potential-energy function for F⃗ ? Let U=0 when x=0.

Answers

  1. Answer:

    The potential energy is -4x^3

    Explanation:

    Given that,

    Force F=-\alpha x^2 i

    We need to calculate the potential energy

    Using formula of work done

    \Delta U=F(x) dx

    Put the value of F

    \Delta U=-\alpha x^2\ dx

    On integration

    U=-\alpha \dfrac{x^3}{3}+C

    U=-\dfrac{\alpha x^3}{3}+C…(I)

    U = 0, x = 0 so C = 0

    Put the value of c and α in equation (I)

    U=-\dfrac{12x^3}{3}+0

    U=-4x^3

    Hence, The potential energy is -4x^3

Leave a Comment