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:
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
\[
\begin{gather}
\bbox[background:#99CCFF; padding:10px; border-radius:10px;]
{K=\frac{mv^2}{2}}
\end{gather}
\]
For the initial speed, the initial kinetic energy will be
\[
\begin{gather}
K_i=\frac{mv_0^2}{2} \\[5pt]
K_i=\frac{(3000\;\mathrm{kg})\times\left(5\;\mathrm{\frac{m}{s}}\right)^2}{2} \\[5pt]
K_i=\frac{(3000\;\mathrm{kg})\times\left(25\;\mathrm{\frac{m^2}{s^2}}\right)}{2}
\end{gather}
\]
\[
\begin{gather}
\bbox[background:#FFCCCC; padding:10px; border-radius:10px;]
{K_i=37500\;\mathrm J=3.75\times 10^{4}\;\mathrm J}
\end{gather}
\]
For the final speed, the final kinetic energy will be
\[
\begin{gather}
K_f=\frac{mv^2}{2} \\[5pt]
K_f=\frac{(3000\;\mathrm{kg})\times\left(20\;\mathrm{\frac{m}{s}}\right)^2}{2} \\[5pt]
K_f=\frac{(3000\;\mathrm{kg})\times\left(400\;\mathrm{\frac{m^2}{s^2}}\right)}{2}
\end{gather}
\]
\[
\begin{gather}
\bbox[background:#FFCCCC; padding:10px; border-radius:10px;]
{K_f=600000\;\mathrm J=6\times 10^{5}\;\mathrm J}
\end{gather}
\]
b) Using the Work-Kinetic Energy Theorem, the work done by force will be the change of kinetic energy
between the two points
\[
\begin{gather}
\bbox[background:#99CCFF; padding:10px; border-radius:10px;]
{W_{\small F}=\Delta K=K_f-K_i}
\end{gather}
\]
\[
\begin{gather}
W_{\small F}=(60\;\mathrm J-3.75\;\mathrm J)\times10^4
\end{gather}
\]
\[
\begin{gather}
\bbox[background:#FFCCCC; padding:10px; border-radius:10px;]
{W_{\small F}=5.6\times 10^5\;\mathrm J}
\end{gather}
\]
c) For the average force exerted by the engine during the acceleration
\[
\begin{gather}
\bbox[background:#99CCFF; padding:10px; border-radius:10px;]
{W_{\small F}=Fd}
\end{gather}
\]
\[
\begin{gather}
F=\frac{W_{\small F}}{d} \\[5pt]
F=\frac{56.25\times 10^4\;\mathrm J}{500\;\mathrm m}
\end{gather}
\]
\[
\begin{gather}
\bbox[background:#FFCCCC; padding:10px; border-radius:10px;]
{F=1125\;\mathrm N}
\end{gather}
\]
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