A 100-g toy car is propelled by a compressed spring that starts it moving. The car follows the curved track in Figure 7.39. Show that the final speed of the toy car is 0.687 m/s if its initial speed is 2.00 m/s and it coasts up the frictionless slope, gaining 0.180 m in altitude.

Conservation of energy: The universe's total energy is conserved. According to conservation of energy, energy neither be created not be destroyed; only the form of energy can be changed.

For a mechanical system, the conservation of energy is given as,

${\text{KE}}_{\text{i}}+{\text{PE}}_{\text{i}}={\text{KE}}_{\text{f}}+{\text{PE}}_{\text{f}}$

Here, ${\text{KE}}_{\text{i}}$is the initial kinetic energy, ${\text{PE}}_{\text{i}}$is the initial potential energy, ${\text{KE}}_{\text{f}}$is the final kinetic energy, and ${\text{PE}}_{\text{f}}$is the final potential energy.

From the conservation of energy,

$\begin{array}{rcl}{\text{KE}}_{\text{i}}+{\text{PE}}_{\text{i}}& =& {\text{KE}}_{\text{f}}+{\text{PE}}_{\text{f}}\\ \frac{1}{2}m{v}_i^2+mg{h}_i& =& \frac{1}{2}m{v}_f^2+mg{h}_f\end{array}$ (1.1)

Here, m is the mass of the toy car $\left(m=100\text{g}\right)$,$v_i$ is the initial speed of the toy car $\left(v_i=2.00\text{m}/\text{s}\right)$, g is the acceleration due to gravity $\left(g=9.8\text{m}/{\text{s}}^2\right)$,$v_f$ is the final speed of the toy car, and $h_f$is the maximum height attained by the toy car on a frictionless slope $\left(h_f=0.180\text{m}\right)$.

Rearranging equation (1.1) in order to get an expression for the final speed of the toy car,

$v_f=\sqrt{v_i^2+2g\left(h_i-h_f\right)}$

Putting all known values,

$\begin{array}{rcl}{v}_f& =& \sqrt{{\left(2.00\text{m}/\text{s}\right)}^2+2\times \left(9.8\text{m}/{\text{s}}^2\right)\times \left[0-\left(0.180\text{m}\right)\right]}\\ & =& 0.687\text{m}/\text{s}\end{array}$

Hence proved, the final speed of the toy car is 0.687m/s.