a 20.0 kg object is at rest, and then speeds up to 16.0 m/s. how much work was done on this system?\n320…

Explanation:

Step1: Recall work - energy theorem

Work done $W=\Delta K$, where $\Delta K$ is change in kinetic energy. Initial kinetic energy $K_i = 0$ (object at rest, $v_i=0$), and final kinetic energy $K_f=\frac{1}{2}mv_f^2$.

Step2: Calculate final kinetic energy

Given $m = 20.0\ kg$ and $v_f=16.0\ m/s$. Using formula $K_f=\frac{1}{2}mv_f^2$, we have $K_f=\frac{1}{2}\times20.0\times(16.0)^2$. $K_f = 10\times256=2560\ J$. Since $W=\Delta K=K_f - K_i$ and $K_i = 0$, $W = K_f$.

Answer:

2,560 J