2. Object 1: weight W_1 = 420 N, a_1 = 9.8 m/s^2 (gravity acceleration on the Earth)

Mass m_1 = W_1 / a_1 = 420/9.8 = 43 kg

Object 2: W_2 = 70 N, a_2 = 1.6 m/s^2, mass m_2 = W_2/a_2 = 70/1.6 = 44 kg

The second object has larger mass, hence it has more inertia than the first one.

3. Given: m = 0.056 kg, v_i = 0, v_f = 39 m/s (initial and final speeds), d = 0.55 m (distance)

Find the force F acting on the object.

The change in kinetic energy K = m(v_f^2 - v_i^2)/2 = 0.056*(39^2 - 0)/2 = 42.6 J.

It equals the work A = F * d done by the force on the distance d.

Answer: F = A/d = 42.6 / 0.55 = 77 N

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