Question

A particle of mass 4.0 kg is constrained to move along the x-axis under a single force F(x) = −cx3 , where c = 8.0 N/m3 . The particle’s speed at A, where xA = 1.0 m, is 6.0 m/s. What is its speed at B, where xB = −2.0 m?

Answer 1

Answer:4.58 m/s

Explanation:

Given

mass of Particle $m=4 kg$

$F=-cx^3$

$a=\frac{F}{m}$

$a=-\frac{cx^3}{m}$

$a=-\frac{8x^3}{4}$

$a=-2x^3$

$v\frac{\mathrm{d} v}{\mathrm{d} x}=-2x^3$

$vdv=-2x^3dx$

integrating

$\int_{6}^{v_b}vdv=\int_{1}^{-2}-2x^3dx$

$\frac{v_b^2-6^2}{2}=-\frac{1}{2}\left [ \left ( -2\right )^4-\left ( 1\right )^4\right ]$

$\frac{v_b^2-36}{2}=-0.5\times 15$

$v_b^2=36-15$

$v_b=\sqrt{21}$

$v_b=4.58 m/s$