Respuesta :
Answer:
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 area bounded by the force function gives the work done by the force.
The speed of the particle at x = 4.0 is approximately 13.66 m/s
Reasons:
The given parameters are;
Particle's mass = 2.0 kg
Force acting on the particle, F(x) = (-5·x² + 7·x) N
The speed of the particle at x = -4.0 m, v = 20.0 m/s
Required:
The speed of the particle at x = 4.0 m
Solution:
Work done by the force, W = F × Δx
Where;
Δx = The change in the position of the particle
Therefore, the work done between -4.0 m, and 4.0 m. is given as follows;
[tex]W = \displaystyle \int\limits^4_{-4} {\left(-5 \cdot x^2 + 7 \cdot x \right)} \, dx = \mathbf{\left[ -\frac{5}{3} \cdot x^3 + \frac{7}{2} \cdot x^2 \right]^4_{-4}}[/tex]
[tex]\displaystyle \left[ -\frac{5}{3} \cdot x^3 + \frac{7}{2} \cdot x^2 \right]^4_{-4} = \left(-\frac{5}{3} \times 4^3 + \frac{7}{2} \times 4^2 \right) - \left(-\frac{5}{3} \times (-4)^3 + \frac{7}{2} \times (-4)^2 \right)[/tex]
[tex]\displaystyle \left(-\frac{5}{3} \times 4^3 + \frac{7}{2} \times 4^2 \right) - \left(-\frac{5}{3} \times (-4)^3 + \frac{7}{2} \times (-4)^2 \right) = -\frac{640}{3} = - 213 .\overline 3[/tex]
Therefore;
The work done on the particle by the force = [tex]\mathbf{-213. \overline 3}[/tex] J
[tex]Kineyic \ enegy, \ K.E. = \mathbf{\dfrac{1}{2} \cdot m \cdot v^2}[/tex]
Initial kinetic energy of the particle, K.E. ar x = 4, is 0.5×2.0 kg×20² = 400 J
According to the work energy theorem, we have;
Work done = Change in kinetic energy
Change in kinetic energy = K.E. at x = 4 - K.E. at x = -4
Therefore;
[tex]-213. \overline 3[/tex] = [tex]\mathbf{\dfrac{1}{2} \times 2 \times v^2}[/tex] - 400
Where;
v = The velocity at x = 4
[tex]\dfrac{1}{2} \times 2 \times v^2 = v^2[/tex]
Which gives;
[tex]-213. \overline 3[/tex] = v² - 400
v² = 400 - [tex]213. \overline 3[/tex] = 186.[tex]\overline 6[/tex]
v = √(186.[tex]\overline 6[/tex]) ≈ 13.66
The speed of the particle at x = 4.0 m is v ≈ 13.66 m/s
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