In an experiment, one of the forces exerted on a proton is F vector = - alpha x^

Question

In an experiment, one of the forces exerted on a proton is F vector = - alpha x^2i, where alpha = 14.8 N/m^2. (a) How much work does F vector do when the proton moves along the straight-line path from the point (0.10 m, 0) to the point (0.10 m, 0.3 m)? 0 J (b) Along the straight-line path from the point (0.12 m, 0) to the point (0.32 m, 0)? J (c) Along the straight-line path from the point (0.32 m, 0) to the point (0.12 m, 0)? J (d) Is the force F vector conservative? Yes No Explain. If F vector is conservative, what is the potential-energy function for it? Let U = 0 when x = 0. (Use the following as necessary: x and alpha. If F vector is not conservative, enter NC.) U =

Explanation / Answer

B) Work done = integral F dx

=integral - ax^2 dx

= - ax^3 /3

= - 14.8*(0.32^3 -0.12^3)

= - 0.4594 J

C) work done = 0.4594 J after applying reverse limit in part b

D) Conservative as we found work done is just negative value of earlier work if initial and final position is just reversed.

E) Potential energy U = integral - Fdx

= integral ax^2

= ax^3 /3

  

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