Question

three bar graphs labeled a, b, c. each has energy (j) on y - axis, categories k (kinetic) and ( u_g ) (gravitational potential) on x - axis. graph a: k bar at 500 j, ( u_g ) bar at 200 j. graph b: k bar at 300 j, ( u_g ) bar at 0 j. graph c: k bar at 500 j, ( u_g ) bar at 0 j. options a, b, c are circular radio buttons.

Explanation:

To solve this, we analyze the kinetic energy (\(K\)) and gravitational potential energy (\(U_g\)) in each graph:

Key Concept:

In a system with mechanical energy conservation (e.g., an object in free fall or moving vertically), the total mechanical energy (\(E = K + U_g\)) remains constant. If an object is at a height where \(U_g = 0\) (e.g., ground level), all energy is kinetic (\(K = E\)) and \(U_g = 0\).

Analyzing Each Option:

• Option A: \(K \approx 500\ \text{J}\), \(U_g \approx 200\ \text{J}\). Total energy \(E = 500 + 200 = 700\ \text{J}\), but \(U_g

eq 0\), so not consistent with a “ground - level” scenario.

• Option B: \(K \approx 300\ \text{J}\), \(U_g = 0\). Total energy \(E = 300\ \text{J}\), but if \(U_g = 0\), \(K\) should equal the total energy (no other forms shown). However, the height of \(K\) here is less than in Option C.
• Option C: \(K \approx 500\ \text{J}\), \(U_g = 0\). This means all mechanical energy is kinetic (e.g., object at ground level, so \(U_g = 0\) and \(K = E\)). This is consistent with a scenario where gravitational potential energy is zero (e.g., object on the ground, moving with kinetic energy equal to total mechanical energy).

Answer:

C