Answer:
v = 5.69 m/s
x = 4.43 m
Explanation:
Mass of the body m = 2.5 kg
force acting on the body [tex]F_x[/tex] = - 7 x N
From the Newton's second Law;
[tex]F_x = ma[/tex]; Then:
[tex]ma = - 7x\\\\a = \frac{-7x}{m}\\\\a = \frac{-7x }{2.5}\\\\a = - 2.8 x\\\\\frac{dv}{dt} = -2.8x\\\\\frac{dv}{dx}*\frac{dx}{dt} = -2.8x\\\\\frac{dv}{dx}v=-2.8 x\\\\vdv = -2.8dx[/tex]
Integrating on both sides ; we have :
[tex]\int\limits {v} \, dv = - 2.8 \int\limits dx \\\\\frac{v^2}{2}= -2.8\frac{x^2}{2}+k\\\\[/tex]
where; k is the integral constant ;
At x = 2.6 m speed is 8.5 m/s
Then;
[tex]\frac{8.5^2}{2.6}= -2.8\frac{2.6}{2}+ k\\\\\frac{72.25}{2.6}= -\frac{7.28}{2}+k\\\\27.79 = -3.64+k\\\\27.79+3.64 = k\\\\k = 31.43[/tex]
However:
[tex]\frac{v^2}{2}= \frac{-2.8x^2}{2}+31.43\\\\\\v^2 = -2.8x^2+62.86 ---- Equation \ 1[/tex]
a)
At x = 3.3 m; speed of the object
[tex]v^2 = -2.8 (3.3)^2 +62.86\\\\v^2 = 32.368\\\\v = \sqrt{32.368}\\\\v = 5.69 \ m/s[/tex]
b)
speed of the body is 2.8 m/s; then
[tex](2.8)62 = -2.8(x)^2 +62.86\\\\2.8x^2 = 62.86 -2.8(x)^2\\\\2.8x^2 = 55.02\\\\x^2 = \frac{55.02}{2.8}\\\\x = \sqrt{19.65}\\\\x = 4.43 \ m[/tex]
