As a 20 kg object moves from 2i5j m to 612jm the constant re

As a 2.0 kg object moves from (2i+5j) m to (61-2j)m, the constant resultant force acting on it is equal to (4i-3j)N. if the speed of the object at the initial positio is 4.0 m/s, what is its kinetic energy at its final position?

Solution

Since the force is (4i - 3j)N, the magnitude here is 5N (sqrt(4^2+3^2))
The magnitude of the acceleration can be determined from F = ma
5N = 2kg * a
a = 2.5 m/s^2

Now find the distance traveled from (2,5) to (6,-2),
x = sqrt((6-2)^2 + (-2-5)^2) = sqrt(65) = 8.06 m

Use the equation 2ax = vf^2 - vi^2, vf is final velocity, vi is initial velocity
2 * (2.5 m/s^2) * (8.06 m) = vf^2 - (4 m/s)^2
vf^2 = 2 * 2.5 * 8.06 + 4^2 = 56.3 m^2/s^2

K.E. = 1/2 mvf^2 = (1/2) * 2kg * 56.3 m^2/s^2 = 56.3 J