1 A car of mass 1000 kg is travelling along a straight, level road.
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{d5a09ed4-a32f-4ff7-aa08-6e54c2ab26a0-1_150_868_316_602}
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\caption{Fig. 6.1}
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- Calculate the acceleration of the car when a resultant force of 2000 N acts on it in the direction of its motion.
How long does it take the car to increase its speed from \(5 \mathrm {~ms} ^ { - 1 }\) to \(12.5 \mathrm {~ms} ^ { - 1 }\) ?
The car has an acceleration of \(1.4 \mathrm {~ms} ^ { - 2 }\) when there is a driving force of 2000 N .
- Show that the resistance to motion of the car is 600 N .
A trailer is now atached to the car, as shown in Fig. 6.2. The car still has a driving force of 2000 N and resistance to motion of 600 N . The trailer has a mass of 800 kg . The tow-bar connecting the car and the trailer is light and horizontal. The car and trailer are accelerating at \(0.7 \mathrm {~ms} ^ { 2 }\).
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{d5a09ed4-a32f-4ff7-aa08-6e54c2ab26a0-1_165_883_1279_554}
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\caption{Fig. 6.2}
\end{figure} - Show that the resistance to the motion of the trailer is 140 N .
- Calculate the force in the tow bar.
The driving force is now removed and a braking force of 610 N is applied to the car. All the resistances to motion remain as before. The trailer has no brakes.
- Calculate the new acceleration. Calculate also the force in the tow-bar, stating whether it is a tension or a thrust (compression).