As a 2.0-kg object moves from (2i+5j)m to (6i-2j)m, the constant resultant force acting on it is equal to (4i-3j)N. If the speed of the object at the initial position is 4.0 m/s, what is its kinetic energy at its final position? Ans: 40J

Answer 1

x distance = 4
y distance = -7

x force = 4
y force = -3

work done on it = 4*4 + -7*-3= 16 + 21 = 37 Joules = increase in Ke
initial Ke = (1/2)(2)(16) = 16 Joules
total energy = 37 + 16 = 53 Joules

Answer 2

your answer is not correct

Answer 3

53 J is the correct answer