91Ó°ÊÓ

i) A single frictionless roller-coaster car of mass$m=825\text{ k²µ}$

ii) Speed of car$\left(v_0\right)=17.0\text{″¾/²õ}$

iii) Height of object from ground$\left(h\right)=42.0\text{″¾}$

iv) Height of point A from ground$\left(h\right)=42.0\text{″¾}$

v) Height of point B from ground

$\left(\frac{h}{2}\right)=21.0\text{″¾}$

The change in potential energy, $\mathit{Δ}\mathit{U}\mathbf{=}\mathbf{−}\mathit{W}$i.e., the gravitational potential energy being equal to the negative of the work done on the object by the gravitational force. Potential energy is defined by the equation.role="math" localid="1661151261384" $\mathit{U}\mathbf{=}\mathit{m}\mathit{g}\mathit{h}$

Work done depends on the initial and final point.

Formula:

$\mathit{U}\mathbf{=}\mathit{m}\mathit{g}\mathit{h}$

Using the formula for potential energy, we can write

$\mathrm{Δ}U=−W$

$\begin{array}{c}W=mgh−mgh\\ W=0\text{ J}\end{array}$

Since the gravitational force is in the vertical direction and displacement is in the horizontal direction,theangle between them is. Hence, the work done is zero, as work done isthedot product of force and displacement.

The displacement between initial point and point B is equal to$\frac{h}{2}$, therefore we can write,

$\begin{array}{c}W=mgh−\frac{mgh}{2}\\ W=\frac{mgh}{2}\end{array}$

Substitute all the value in the above equation.

$\begin{array}{c}W=\frac{825\text{ k²µ}×9.8\text{″¾}/\text{s}×42.0\text{″¾}}{2}\\ =169785\text{ J}\\ =1.70×{10}^5\text{ J}\end{array}$

Hence the work is,$1.70×{10}^5\text{ J}$.

Displacement between initial point and final point is$h$, and it is along the direction of the gravitational force.

$W=mgh$

Substitute all the value in the above equation.

$\begin{array}{c}W=825\text{ k²µ}×9.8\text{″¾}/\text{s}×42.0\text{″¾}\\ =339570\text{ J}\\ =3.40×{10}^5\text{ J}\end{array}$

Hence the work is,$3.40×{10}^5\text{ J}$

If the gravitational potential of the car at point C taken to be zero, then at point B, we have height as$\frac{h}{2}$, so we can write,

$\begin{array}{l}{U}_B=\frac{mgh}{2}−0\\ U_B=\frac{mgh}{2}\end{array}$

Substitute all the value in the above equation.

$\begin{array}{c}{U}_B=\frac{825\text{ k²µ}×9.8\text{″¾}/\text{s}×42.0\text{″¾}}{2}\\ =169785\text{ J}\\ =1.70×{10}^5\text{ J}\end{array}$

Hence the potential energy is,$1.70×{10}^5\text{ J}$

Similarly, for point A, the height is$h$, so we can write.

$\begin{array}{l}W=mgh−0\\ W=mgh\end{array}$

Substitute all the value in the above equation.

$\begin{array}{c}W=825\text{ k²µ}×9.8\text{″¾}/\text{s}×42.0\text{″¾}\\ =339570\text{ J}\\ =3.40×{10}^5\text{ J}\end{array}$

Hence the work is,$3.40×{10}^5\text{ J}$

If mass were doubled, then potential energy is also doubled because potential energy is directlyproportional to mass.