A particle of mass 2.0 kg moves under the influence of the force F(x)=(-5x^2+7x)~\text{N}F(x)=(−5x 2 +7x) N. If its speed at x=-4.0~\text{m}x=−4.0 m is v=20.0~\text{m/s}v=20.0 m/s, what is its speed at x=4.0~\text{m}x=4.0 m?

The speed of the particle at x = 4.0 m is 13.66 m/s

Explanation:

The work done by this force between the two points above is given by

W = ∫ F dx

W = ∫⁴₋₄ (-5x² + 7x) dx

W = [(-5x³/3) + (7x²/2)]⁴₋₄

W = [(-5(4³)/3) + (7(4²)/2)] - [(-5(-4)³)/3) + (7((-4)²)/2)] = (-50.6667) - (162.6667) = (- 213.33 J)

Kinetic energy at -4.0 m

At this point, v = 20 m/s

K.E = mv²/2 = 2 × 20²/2 = 400 J

To obtain the kinetic energy at 4 m,

We apply the work-energy theorem which mathematically translates to

The work done in moving a particle from one point to another = Change in kinetic energy of the particle between those two points

W = ΔK.E

Work done between x = - 4m and x = 4 m is - 213.33 J

Hence, ΔK.E = -213.33 J

Change in kinetic energy of the particle between x = - 4m and x = 4m is ΔK.E

ΔK.E = (Kinetic energy at x = 4m) - (kinetic energy at x = - 4m)

- 213.33 J = (mv²/2) - 400

mv²/2 = -213.33 + 400 = 186.67 J

2v² = 2 × 186.67

v² = 186.67

v = 13.66 m/s.

the positions which points up in upward carbon and down in downward carbon are axial

and vice versa holds true for equatorial positions

so the correct answer with labelled diagram will be