Question

As a 2.0-kg object moves from (4.4 i + 5j) m to ( 11.6 i - 2j) m, the constant resultant force

acting on it is equal to (41 - 9j) 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?
Answer:

Answer 1

v_f = 10.38 m / s

Step-by-step explanation:

For this exercise we can use the relationship between work and kinetic energy

W = ΔK

note that the two quantities are scalars

Work is defined by the relation

W = F. Δx

the bold are vectors. The displacement is

Δx = r_f -r₀

Δx = (11.6 i - 2j) - (4.4 i + 5j)

Δx = (7.2 i - 7 j) m

W = (4 i - 9j). (7.2 i - 7 j)

remember that the dot product

i.i = j.j = 1

i.j = 0

W = 4 7.2 + 9 7

W = 91.8 J

the initial kinetic energy is

Ko = ½ m vo²

Ko = ½ 2.0 4.0²

Ko = 16 J

we substitute in the initial equation

W = K_f - K₀

K_f = W + K₀

½ m v_f² = W + K₀

v_f² = 2 / m (W + K₀)

v_f² = 2/2 (91.8 + 16)

v_f = √107.8

v_f = 10.38 m / s