Question

In Figure 8-29, a single frictionless roller-coaster car of mass$\mathit{m}\mathbf{}\mathbf{=}\mathbf{}\mathbf{825}\mathbf{\text{\hspace{0.17em}kg}}$tops the first hill with speed ${\mathit{v}}_{\mathbf{0}}\mathbf{}\mathbf{}\mathbf{=}\mathbf{}\mathbf{17}\mathbf{.0}\mathbf{}\mathbf{\text{\hspace{0.17em}m/s}}$at height$\mathit{h}\mathbf{}\mathbf{=}\mathbf{}\mathbf{42}\mathbf{.0}\mathbf{\text{\hspace{0.17em}m}}$.How much work does the gravitational force do on the car from that point to (a) Point A, (b) Point B, (c) Point C? If the gravitational potential energy of the car-Earth system is taken to be zero at C, what is its value when the car is at (d) B, (e) A? (f) If mass m were doubled, would the change in the gravitational potential energy of the system betweenpoints A and B increase, decrease, or remain the same?

Short answer

a) Work done form initial point to point$A=0\text{\hspace{0.17em}J}$

b) Work done form initial point to point$B=1.70\times {10}^5\text{\hspace{0.17em}J}$

c) Work done form initial point to point$C=3.40\times {10}^5\text{\hspace{0.17em}J}$

If the gravitational Potential of the car at point C taken to be zero

d) Work done at point$B=1.70\times {10}^5\text{\hspace{0.17em}J}$

e) Work done at point$A=3.40\times {10}^5\text{\hspace{0.17em}J}$

f) If Mass mwere doubled, then potential energy between point A and B also doubled i.e. total potential energy increases.

Step-by-step solution

Given

i) A single frictionless roller-coaster car of mass$m=825\text{\hspace{0.17em}kg}$

ii) Speed of car$\left(v_0\right)=17.0\text{\hspace{0.17em}m/s}$

iii) Height of object from ground$\left(h\right)=42.0\text{\hspace{0.17em}m}$

iv) Height of point A from ground$\left(h\right)=42.0\text{\hspace{0.17em}m}$

v) Height of point B from ground

$\left(\frac{h}{2}\right)=21.0\text{\hspace{0.17em}m}$

vi) Height of point from ground$=0$

 Step 2: To understand the concept

The change in potential energy, $\mathit{\Delta }\mathit{U}\mathbf{=}\mathbf{-}\mathit{W}$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:

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

(a) Calculate much work does the gravitational force do on the car from that point-to-point A

Using the formula for potential energy, we can write

$\Delta U=-W$

$\begin{array}{c}W=mgh-mgh\\ W=0\text{\hspace{0.17em}J}\end{array}$

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.

(b) Calculate much work does the gravitational force do on the car from that point-to-point B

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.

$\begin{array}{c}W=\frac{825\text{\hspace{0.17em}kg}\times 9.8\text{\hspace{0.17em}m}/\text{s}\times 42.0\text{\hspace{0.17em}m}}{2}\\ =169785\text{\hspace{0.17em}J}\\ =1.70\times {10}^5\text{\hspace{0.17em}J}\end{array}$

Hence the work is,$1.70\times {10}^5\text{\hspace{0.17em}J}$.

(c) Calculate much work does the gravitational force do on the car from that point-to-point C

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.

$\begin{array}{c}W=825\text{\hspace{0.17em}kg}\times 9.8\text{\hspace{0.17em}m}/\text{s}\times 42.0\text{\hspace{0.17em}m}\\ =339570\text{\hspace{0.17em}J}\\ =3.40\times {10}^5\text{\hspace{0.17em}J}\end{array}$

Hence the work is,$3.40\times {10}^5\text{\hspace{0.17em}J}$

(d) Calculate the gravitational potential energy of the car-Earth system at B if it is taken to be zero at C

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.

$\begin{array}{c}{U}_B=\frac{825\text{\hspace{0.17em}kg}\times 9.8\text{\hspace{0.17em}m}/\text{s}\times 42.0\text{\hspace{0.17em}m}}{2}\\ =169785\text{\hspace{0.17em}J}\\ =1.70\times {10}^5\text{\hspace{0.17em}J}\end{array}$

Hence the potential energy is,$1.70\times {10}^5\text{\hspace{0.17em}J}$

(e) Calculate the gravitational potential energy of the car-Earth system at A if it is taken to be zero at C 

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.

$\begin{array}{c}W=825\text{\hspace{0.17em}kg}\times 9.8\text{\hspace{0.17em}m}/\text{s}\times 42.0\text{\hspace{0.17em}m}\\ =339570\text{\hspace{0.17em}J}\\ =3.40\times {10}^5\text{\hspace{0.17em}J}\end{array}$

Hence the work is,$3.40\times {10}^5\text{\hspace{0.17em}J}$

(f) If mass m were doubled, find out if the gravitational potential energy of the system between points A and B increase, decrease, or remain the same

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