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Feb/Mar 2021 p42 q2
3410
A car of mass 1400 kg is travelling at constant speed up a straight hill inclined at \(\alpha\) to the horizontal, where \(\sin \alpha = 0.1\). There is a constant resistance force of magnitude 600 N. The power of the car’s engine is 22 500 W.
(a) Show that the speed of the car is 11.25 m s\(^{-1}\).
The car, moving with speed 11.25 m s\(^{-1}\), comes to a section of the hill which is inclined at 2° to the horizontal.
(b) Given that the power and resistance force do not change, find the initial acceleration of the car up this section of the hill.
Solution
(a) The driving force \(DF\) is given by \(DF = \frac{22500}{v}\). Applying Newton's second law for constant speed (\(a = 0\)), we have:
\(DF - 1400g \times 0.1 - 600 = 0\)
Solving for \(v\), we find \(v = 11.25 \text{ m s}^{-1}\).
(b) For the section inclined at 2°, using Newton's second law:
\(DF - 1400g \sin 2^{\circ} - 600 = 1400a\)
Substitute \(DF = \frac{22500}{11.25}\) and solve for \(a\):