Solved Problem on Work and Energy

A car with a mass of 3000 kg accelerates under the action of a constant force from an initial speed of 18 km/h to 72 k /h. Determine:
a) The initial and final kinetic energies of the car;
b) What is the work done by the force between the initial and final instants?
c) What is the force needed to accelerate the car on a 500 m route?

Problem data:

Mass of the car: m = 3000 kg;
Initial speed of the car: v₀ = 18 km/h;
Final speed of the car: v = 72 km/h.

Problem diagram:

Figure 1

Solution

First, we convert the speeds given in kilometers per hour (km/h) to meters per second (m/s) used in the International System of Units (SI)

\[ \begin{gather} v_0=18\;\frac{\cancel{\mathrm{km}}}{\mathrm{\cancel h}}\times\frac{1000\;\mathrm m}{1\;\mathrm{\cancel{km}}}\frac{1\;\mathrm{\cancel{h}}}{3600\;\mathrm s}=5\;\mathrm{m/s}\\[10pt] v=72\;\frac{\cancel{\mathrm{km}}}{\mathrm{\cancel{h}}}\times\frac{1000\;\mathrm m}{1\;\mathrm{\cancel{km}}}\frac{1\;\mathrm{\cancel{h}}}{3600\;\mathrm s}=20\;\mathrm{m/s} \end{gather} \]

a) The kinetic energy is given by

\[ \begin{gather} \bbox[#99CCFF,10px] {K=\frac{mv^2}{2}} \end{gather} \]

For the initial speed, the initial kinetic energy will be

\[ \begin{gather} K_{i}=\frac{mv_0^2}{2}\\[5pt] K_{i}=\frac{(3000\;\mathrm{kg})\times\left(5\;\mathrm{\frac{m}{s}}\right)^2}{2}\\[5pt] K_{i}=\frac{(3000\;\mathrm{kg})\times\left(25\;\mathrm{\frac{m^2}{s^2}}\right)}{2} \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {K_{i}=37500\;\mathrm J=3.75\times 10^{4}\;\mathrm J} \end{gather} \]

For the final speed, the final kinetic energy will be

\[ \begin{gather} K_{f}=\frac{mv^2}{2}\\[5pt] K_{f}=\frac{(3000\;\mathrm{kg})\times\left(20\;\mathrm{\frac{m}{s}}\right)^2}{2}\\[5pt] K_{f}=\frac{(3000\;\mathrm{kg})\times\left(400\;\mathrm{\frac{m^2}{s^2}}\right)}{2} \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {K_{f}=600000\;\mathrm J=6\times 10^{5}\;\mathrm J} \end{gather} \]

b) Using the Work-Kinetic Energy Theorem, the work done by force will be the change of kinetic energy between the two points

\[ \begin{gather} \bbox[#99CCFF,10px] {W_{\small F}=\Delta K=K_f-K_i} \end{gather} \]

\[ \begin{gather} W_{\small F}=(60\;\mathrm J-3.75\;\mathrm J)\times10^4 \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {W_{\small F}=5.6\times 10^5\;\mathrm J} \end{gather} \]

c) For the average force exerted by the engine during the acceleration

\[ \begin{gather} \bbox[#99CCFF,10px] {W_{\small F}=Fd} \end{gather} \]

\[ \begin{gather} F=\frac{W_{\small F}}{d}\\[5pt] F=\frac{56.25\times 10^4\;\mathrm J}{500\;\mathrm m} \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {F=1125\;\mathrm N} \end{gather} \]

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