Let's solve the problem shown in the image!

Given:

• Mass of the cart (m) = 200 kg
• Radius of the loop (R) = 5.00 m
• Initial height $h = 4R$
• Height at the top of the loop (point A) = $2R$
• Gravitational acceleration (g) = 9.81 m/s²

We are asked for the work done by gravitational force on the cart as it moves from the initial height $h$ to the top of the loop (point A).

1. Concept:

The work done by the gravitational force can be found by calculating the change in gravitational potential energy:

$W_{\text{gravity}} = U_{\text{initial}} - U_{\text{final}}$

where the gravitational potential energy is given by:

$U = m g h$

2. Calculating Initial and Final Potential Energy:

• Initial height: $h_{\text{initial}} = 4R = 4(5) = 20 \, \text{m}$
  So,
  $U_{\text{initial}} = m g h_{\text{initial}} = (200)(9.81)(20) = 39,240 \, \text{J}$

• Final height at point A: $h_{\text{final}} = 2R = 2(5) = 10 \, \text{m}$
  So,
  $U_{\text{final}} = m g h_{\text{final}} = (200)(9.81)(10) = 19,620 \, \text{J}$

3. Work Done by Gravity:

$W_{\text{gravity}} = U_{\text{initial}} - U_{\text{final}}$ $W_{\text{gravity}} = 39,240 - 19,620 = 19,620 \, \text{J}$

4. Answer:

The work done by the gravitational force is:

$\boxed{19,600 \, \text{J}}$

Would you like any further clarification or additional steps explained?

Here are 5 related questions for further practice:

1. What is the total mechanical energy at the top of the loop?
2. How would the answer change if the radius of the loop were doubled?
3. What is the velocity of the cart at the top of the loop?
4. How does friction affect the energy conservation in this problem?
5. What forces act on the cart at the top of the loop?

Tip: Always double-check units to ensure accuracy in physics problems.