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.

i believe the answer is true.