Question

questions 2 - 8: frictionless rollercoaster
fill in the tables below to calculate the kinetic energy of the cart at point 2 for the following values: ( m = 2.000 , \text{kg}, h_i = 30 , \text{m}, h_f = 10 , \text{m} )

	point 1 (initial)		point 2 (final)
( e_p , (\text{j}) )			( e_p , (\text{j}) )
( e_k , (\text{j}) )			( e_k , (\text{j}) )
( e_{\text{total}} , (\text{j}) )			( e_{\text{total}} , (\text{j}) )

Explanation:

Step1: Calculate Potential Energy at Point 1 ($E_{p1}$)

The formula for gravitational potential energy is $E_p = mgh$, where $m = 2.000\ kg$, $g = 9.8\ m/s^2$, and $h_1 = 30\ m$.
$E_{p1} = 2.000 \times 9.8 \times 30 = 588\ J$

Step2: Calculate Kinetic Energy at Point 1 ($E_{k1}$)

Since the velocity at point 1 ($v_1$) is 0, the kinetic energy formula $E_k = \frac{1}{2}mv^2$ gives $E_{k1} = \frac{1}{2} \times 2.000 \times 0^2 = 0\ J$

Step3: Calculate Total Energy at Point 1 ($E_{Total1}$)

Total energy is the sum of potential and kinetic energy: $E_{Total1} = E_{p1} + E_{k1} = 588 + 0 = 588\ J$

Step4: Calculate Potential Energy at Point 2 ($E_{p2}$)

Using $E_p = mgh$ with $h_2 = 10\ m$: $E_{p2} = 2.000 \times 9.8 \times 10 = 196\ J$

Step5: Calculate Total Energy at Point 2 ($E_{Total2}$)

In a frictionless system, total mechanical energy is conserved, so $E_{Total2} = E_{Total1} = 588\ J$

Step6: Calculate Kinetic Energy at Point 2 ($E_{k2}$)

Using $E_{Total2} = E_{p2} + E_{k2}$, we solve for $E_{k2}$: $E_{k2} = E_{Total2} - E_{p2} = 588 - 196 = 392\ J$

Filling the Table:

Point 1 (initial)

• $E_p (J)$: $588$
• $E_k (J)$: $0$
• $E_{Total} (J)$: $588$

Point 2 (final)

• $E_p (J)$: $196$
• $E_k (J)$: $392$
• $E_{Total} (J)$: $588$

Answer:

Point 1 (initial)

• $E_p (J)$: $\boldsymbol{588}$
• $E_k (J)$: $\boldsymbol{0}$
• $E_{Total} (J)$: $\boldsymbol{588}$

Point 2 (final)

• $E_p (J)$: $\boldsymbol{196}$
• $E_k (J)$: $\boldsymbol{392}$
• $E_{Total} (J)$: $\boldsymbol{588}$