## Related questions with answers

A 100-g toy car is propelled by a compressed spring that starts it moving. The car follows the curved track. 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.

Solution

VerifiedHere the mass of the toy car

$\begin{align*} m = 100\,\mathrm{g} = 0.100\,\mathrm{kg} \end{align*}$

Initial speed of the car

$\begin{align*} v_{\text{i}} = 2.00\,\mathrm{m\,s^{-1}} \end{align*}$

By taking the ground as reference, the initial height of the car

$\begin{align*} h_{\text{i}} = 0.00\,\mathrm{m} \end{align*}$

And the final height of the car

$\begin{align*} h_{\text{f}} = 0.180\,\mathrm{m} \end{align*}$

Now according to the law of conservation of energy, we get

$\begin{align*} KE_{\text{i}} + PE_{\text{i}} = KE_{\text{f}} + PE_{\text{f}} \end{align*}$

OR

$\begin{align*} \frac{1}{2}\,m\,v_{\text{i}}^2 + m\,g\,h_{\text{i}} = \frac{1}{2}\,m\,v_{\text{f}}^2 + m\,g\,h_{\text{f}} \end{align*}$

OR

$\begin{align*} v_{\text{f}} & = \sqrt{v_{\text{i}}^2 + 2\,g\,h_{\text{i}} - 2\,g\,h_{\text{f}}}\\ & = \sqrt{v_{\text{i}}^2 + 2\,g\,\times 0 - 2\,g\,h_{\text{f}}}\\ & = \sqrt{\left(2.00\,\mathrm{m\,s^{-1}} \right)^2 - 2\,\times 9.80\,\mathrm{m\,s^{-2}}\,\times 0.180\,\mathrm{m}}\\ & = 0.687\,\mathrm{m\,s^{-1}} \end{align*}$

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