Answer to Question #350253 in Physics for mlas2

Question

A 1450-kg

 car is traveling with an initial speed of 24.0 m/s.

Determine the car's speed after −1.90 ✕ 10⁵ J

 of net work is done on the car.

 

Question 5.2b:

A 11.8-kg

 box is sliding across a horizontal floor. It has an initial speed of 1.45 m/s

 and the only force acting on it is kinetic friction with magnitude f_k = 2.65 N.

Determine the distance the box will travel before coming to rest.

Answer 1

1. The energy-work theorem says

$$\frac{mv_f^2}{2}=\frac{mv_i^2}{2}+W$$

$$v_f=\sqrt{v_i^2+2W/m}$$

$$v_f=\sqrt{24.0^2+2*(-1.90*10^5)/1450}=17.7\:\rm m/s$$

2. The energy-work theorem says

$$\frac{mv_f^2}{2}-\frac{mv_i^2}{2}=W$$

$$0-\frac{mv_i^2}{2}=-f_kd$$

So

$$d=\frac{mv_i^2}{2f_k}=\frac{11.8*1.45^2}{2*2.65}=4.68\:\rm m$$