Question

A police officer in hot pursuit drives her car through a circular turn of radius 379 mwith a constant speed of 90.0 km/h. Her mass is 52.0 kg.What are (a) themagnitude and (b) the angle (relative to vertical) of the net force of the officer onthe car seat? (Hint: Consider both horizontal and vertical forces.)

Answer 1

The centripetal force keeps the car in the circle. This force is directed towards the center of the circle. The upward force on the police officer is f = mg and the centripetal force is represented as follows

`F_c=(mv^2)/(r)`

The net force will be

`\begin{gathered} F_(net)=F_c+mg \\ F_(net)=\sqrt[]{((mv^2)/(r))^2+(mg)^2} \end{gathered}`

A.

`\begin{gathered} r=379m \\ m=52.0\text{ kg} \\ \text{speed}=90\text{ km/h} \\ \text{speed}=25\text{ m/s} \\ F_(net)=\sqrt[]{((52*25^2)/(379))^2+(52*9.8)^2} \\ F_(net)=\sqrt[]{((32500)/(379))^2+(509.6)^2} \\ F_(net)=\sqrt[]{(85.7519788918)^2+259692.16} \\ F_(net)=\sqrt[]{7353.40188386+259692.16} \\ F_(net)=\sqrt[]{267045.561884} \\ F_(net)=516.764512988 \\ F_(net)\approx516.76N \end{gathered}`

B.

`\begin{gathered} \tan \emptyset=((mv^2)/(r))/(mg) \\ \tan \emptyset=(mv^2)/(r)*(1)/(mg) \\ \tan \emptyset=(v^2)/(rg) \\ \tan \emptyset=(25^2)/(379*9.8) \\ \tan \emptyset=(625)/(3714.2) \\ \tan \emptyset=0.16827311399 \\ \emptyset=\tan ^(-1)0.16827311399 \\ \emptyset=9.55185382438 \\ \emptyset=9.55^(\circ) \end{gathered}`