Question

A $150$ g particle at $x=0$ is moving at $2.00$ m/s in the $+x-$direction. As it moves, it experiences a force given by $F_x=(0.250N)\sin (x/2.00m)$. What is the particle’s speed when it reaches $x=3.14$ m?

Short answer

Velocity of the particle at given $x$is $3.26$m/s

Step-by-step solution

Step 1. Given Information

Mass of paricle$m=150g=0.15kg$

Initial Velocity $v_i=2$m/s

Initial position $x_i=0$m

Final position $x_f=3.14$m

Variable force acting$F_x=(0.250N)\sin (x/2m)$

Final velocity$v_f=?$

Step 2. Find the velocity of a particle

We have given variable force acting on a particle and we need to find the velocity at given $x$. We can do so by calculating work done by the force for moving particles from $x=0$to $x=3.14$m and then by equating total work done as a change in kinetic energy

$$\begin{gathered} W={\int }_{x_i}^{x_f}{F}_{x}dx \\ ={\int }_0^{3.14}0.250\sin \frac{x}{2}dx \\ =-0.50\cos \frac{x}{2} \\ =0.5J \end{gathered}$$

Now,

$$\begin{gathered} W=∆K \\ W=\frac{1}{2}m{v_f}^2-\frac{1}{2}m{v_i}^2 \\ \frac{1}{2}m{v_f}^2=W+\frac{1}{2}m{v_i}^2 \\ =0.5+\frac{1}{2}\times 0.15\times 2^2=0.8 \\ {v_f}^2=\frac{2\times 0.8}{0.15}=10.67 \\ {v}_f=3.26m/s \end{gathered}$$