Question
A car of mass \(1500\,\mathrm{kg}\) is travelling along a straight horizontal road at constant velocity \(v\). The car is subject to a total resistive force \(F\), as shown in Fig. 3.1.
(a) Show that the power \(P\) developed by the engine in overcoming the total resistive force is given by the equation
\(P=Fv\) (2 marks)
(b) The car now moves up a slope at a constant speed of \(30\,\mathrm{m\,s^{-1}}\). The slope is at an angle of \(6.0^\circ\) to the horizontal as shown in Fig. 3.2.
The total resistive force acting on the car is \(1600\,\mathrm{N}\).
(i) Show that the increase in gravitational potential energy of the car in a time of \(1.0\,\mathrm{s}\) is \(46000\,\mathrm{J}\). (2 marks)
________________________________________________________________________________
(ii) Use the information in (b)(i) to determine the power developed by the engine to move the car up the slope. (2 marks)
power = ______________________________ \(\mathrm{W}\)
(c) The car picks up a passenger and then continues up the slope at the same speed as in (b).
(i) State and explain the effect, if any, that the passenger has on the air resistance acting on the car. (1 mark)
_______________________________________________________
(ii) State and explain the effect, if any, that the passenger has on the power developed by the engine. (1 mark)
________________________________________________________
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702):
• 5.2: Gravitational potential energy and kinetic energy — part (b)(i)
• 5.3: Power — parts (a), (b)(ii) and (c)(ii)
▶️ Answer/Explanation
(a) Power developed by the engine [2 marks]
The work done in overcoming a force \(F\) through a distance \(d\) is
\(W=Fd\)
Power is the rate of doing work:
\(P=\dfrac{W}{t}\)
Therefore,
\(P=\dfrac{Fd}{t}\)
Since \(v=\dfrac{d}{t}\),
\(P=Fv\)
Answer: \( \boxed{P=Fv} \)
(b)(i) Increase in gravitational potential energy [2 marks]
In \(1.0\,\mathrm{s}\), the car travels \(30\,\mathrm{m}\) along the slope.
The vertical height gained is
\(\Delta h=30\sin6.0^\circ\)
The increase in gravitational potential energy is
\(\Delta E_{\mathrm{p}}=mg\Delta h\)
\(\Delta E_{\mathrm{p}}=1500\times9.81\times30\sin6.0^\circ\)
\(\Delta E_{\mathrm{p}}\approx4.6\times10^4\,\mathrm{J}\)
Answer: \( \boxed{46000\,\mathrm{J}} \)
(b)(ii) Power developed by the engine [2 marks]
The engine must provide power to increase the gravitational potential energy and to overcome the resistive force.
Power required to overcome the resistive force is
\(P_{\mathrm{resistive}}=Fv\)
\(P_{\mathrm{resistive}}=1600\times30\)
\(P_{\mathrm{resistive}}=48000\,\mathrm{W}\)
The power required to increase gravitational potential energy is
\(P_{\mathrm{GPE}}=\dfrac{46000}{1.0}=46000\,\mathrm{W}\)
Therefore,
\(P=48000+46000\)
\(P=94000\,\mathrm{W}\)
Answer: \( \boxed{9.4\times10^4\,\mathrm{W}} \)
(c)(i) Effect on air resistance [1 mark]
The speed of the car remains the same. Therefore, the air resistance remains the same.
Answer: \( \boxed{\text{Air resistance is unchanged.}} \)
(c)(ii) Effect on power developed by the engine [1 mark]
The passenger increases the total mass of the car.
Therefore, the component of the car’s weight acting down the slope increases. The engine must provide more power to move the heavier car up the slope at the same speed.
Answer: \( \boxed{\text{The power developed by the engine increases.}} \)