When a 2 kg object is at x = 2 m it has a velocity 2.5 m/s in the positive x-direction.The object is subjected to a single force of the form F=3x² + 2x + 5 N as it moves from x = 2 m to x = 4 m.what is the speed of the object when its location is x = 4 m?

Question

Quizplus · Accepted Answer

Work done by the force on the object is W=∫24(3x²+2x+5)dx=[x³+x²+5x]24=103 JBy the work-energy theorem,W=ΔK=Kf−Ki103=Kf−(1/2)(2 kg)(2.5 m/s)²Kf=128.75 JKf=(1/2)mvf²vf=9.2 m/s

When a 2 Kg Object Is at X = 2