Answer
See the detailed answer.
Work Step by Step
To find the work done on our object in each stage, we need to use the work-kinetic energy theorem which is given by
$$W=\Delta K= \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$$
$$W= \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$$
So,
- In the first stage, from A to B, $v_i=v_A=2$ m/s, and $v_f=v_B=-2$ m/s.
Thus,
$$W_1= \frac{1}{2}m(v_B^2 - v_A^2)=\frac{1}{2}(2)[(-2)^2 - (2)^2]$$
$$W_1= \color{red}{\bf 0}\;\rm J$$
- In the second stage, from B to C, $v_i=v_B=-2$ m/s, and $v_f=v_C=-2$ m/s.
Thus,
$$W_2= \frac{1}{2}m(v_C^2 - v_B^2)=\frac{1}{2}(2)[(-2)^2 - (-2)^2]$$
$$W_2= \color{red}{\bf 0}\;\rm J$$
- In the third stage, from C to D, $v_i=v_C=-2$ m/s, and $v_f=v_D=0$ m/s.
Thus,
$$W_3= \frac{1}{2}m(v_D^2 - v_C^2)=\frac{1}{2}(2)[(0)^2 - (-2)^2]$$
$$W_3= \color{red}{\bf -4}\;\rm J$$
- In the fourth stage, from D to E, $v_i=v_D=0$ m/s, and $v_f=v_E=2$ m/s.
Thus,
$$W_4= \frac{1}{2}m(v_E^2 - v_D^2)=\frac{1}{2}(2)[(2)^2 - (0)^2]$$
$$W_4= \color{red}{\bf 4}\;\rm J$$
