Answer:
6.57 m/s
Explanation:
From work energy theorem, work done = W = Change in Kinetic energy
W = [tex]\int F dx = \frac{1}{2} m v^{2}- \frac{1}{2} m u^{2}[/tex]
here v is the final velocity and u is the initial velocity and F is the force.
Horizontal component of the force is considered here since motion is along the horizontal (X) direction. So cos 31
Work done =[tex]\int_{1}^{1.5}(5.00N/m^{2})x^{2} cos 31 dx[/tex]
⇒ [tex](5)(0.866)[\frac{1.5^{3}}{3}-\frac{1^{3}}{3}][/tex] =3.39 J
[tex]\frac{1}{2} m v^{2}- \frac{1}{2} m u^{2}[/tex] = 3.39
⇒ [tex]\frac{1}{2} (0.25) v^{2}- \frac{1}{2} (0.25)4^{2}[/tex] = 3.30
⇒ Speed at x = 1.50 m = v = 6.57 m/s
