A 0.333 kg particle with an initial velocity of 1.89 m/s is accelerated by a constant force of 1.64 N over a distance of 0.281 m. Use the concept of energy to determine the final velocity of the particle. (It is useful to double-check your answer by also solving the problem using Newton's Laws and the kinematic equations.)

2.5 m/s

Explanation:

m = 0.333 kg

u = 1.89 m/s

F = 1.64 N

S = 0.281 m

Let the final velocity is v.

By energy conservation

Work done = Change in kinetic energy

F x S = 1/2 m (v^2 - u^2)

1.64 x 0.281 = 0.5 x 0.333 (v^2 - 1.89^2)

2.767 = v^2 - 3.5721

v^2 = 6.339

v = 2.5 m/s

By using kinematics

Let a be the acceleration

a = F / m = 1.64 / 0.333 = 4.925 m/s^2

Use third equation of motion

v^2 = u^2 + 2as

v^2 = 1.89^2 + 2 x 4.925 x 0.281

v^2 = 6.339

v = 2.5 m/s