Conservation of Energy / Work problem (AP Physics C: Mechanics)

Question

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 

F_x = (0.750N)sin(x/2.00m).

What is the particle's speed when it reaches x=3.14m?

Answer 1

I will set this up for you, but YOU do the calculations.

Find the work done by the force. That will give the change in kinetic energy. You have enough information to get the initial kinetic energy. From these two quantities, you can calculate the final speed.

Work = change in kinetic energy

∫F(x)dx = K₂ - K₁

∫₀^3.14 0.750sin(x/2.00)dx = m(v₂² - v₁²)/2

You know m and v₁. You want to find v₂.

-0.750(2.00)[cos(3.14/2.00) - cos(0)] = 0.150(v₂² - 8.00²)/2

From here on, just do the algebra to solve for v₂, and you are done. Note I converted the mas to kg. The result will be in m/s. Be sure you are working in radians.