Answer to Question #115191 in Physics for yamann

Question

A 73 kg object slowed down to 3.0 m/s. If the work done on this object is -2.0 x 10^3 J, then what is the initial speed

Answer 1

According to the kinetic energy theorem, the work was spent to slow the body down:

$KE_f-KE_i=W,\\ \frac{1}{2}mu^2-\frac{1}{2}mv^2=W,\\ \space\\ \frac{1}{2}mv^2=\frac{1}{2}mu^2-W,\\ \space\\ v=\sqrt{u^2-\frac{2W}{m}}=\sqrt{3^2-\frac{2(-2.0 \cdot 10^3)}{73}}=8\text{ m/s}.$