Question

A 150 g particle at x = 0 is moving at 8.00 m/s in the +x-direction....

A 150 g particle at x = 0 is moving at 8.00 m/s in the +x-direction. As it moves, it experiences a force given by Fx=(0.350N)sin(x/2.00m).

What is the particles speed when it reaches x = 3.14 m?

Express your answer with the appropriate units.

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Answer #1

vo = initial speed at x = 0 , = 8 m/s

v = speed at x = 3.14 m , = ?

m = mass of particle = 150 g = 0.150 kg

Fx = force acting on the particle = (0.350) Sin(x/2)

Using work-energy theorem

Fx dx = (0.5) m (v2 - vo2)

(0.350) Sin(x/2) dx = (0.5) m (v2 - vo2)

(0.350) (1.99841) = (0.5) (0.150) (v2 - 82)

v = 8.6 m/s

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