Solved Energy Bar Charts 1) Complete the empty energy bar | Chegg.com - Free Printable

Let's solve each part of this Energy Bar Chart problem step by step. We are analyzing the energy transformations in a system involving a cart (or block) moving along a track, with different conditions: no friction vs. with friction, and initial height considerations.

We will consider:

- K: Kinetic Energy
- U_g: Gravitational Potential Energy
- U_s: Elastic Potential Energy (spring, if any)
- E_tot: Total Mechanical Energy = K + U_g + U_s

In all cases, we assume:
- The system is the cart + Earth (so gravity is internal).
- If there’s no friction, mechanical energy is conserved.
- If there is friction, some mechanical energy is lost to thermal energy (not shown on bar charts unless specified).

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a) No friction, not moving at C

Description:
A cart starts at position A (on a flat surface), goes up an incline to point B, then reaches point C at the top where it stops (not moving). There is no friction.

#### Step-by-step Analysis:

- Position A:
- Cart is at rest → K = 0
- At lowest point → U_g = 0 (we can set reference here)
- So total energy: E_tot = 0 + 0 = 0? Wait — but that can’t be right because it moves upward.

Wait! Let's look again. In part a), the diagram shows the cart starting at A, going up to B, then to C, which is higher than B.

But the energy bar chart at A already shows a non-zero kinetic energy (K > 0), and U_g = 0, so E_tot = K.

So the reference point for gravitational potential energy (U_g) must be at A, so U_g(A) = 0.

At position C, the cart is not moving → K = 0
And it's at the highest point → U_g is maximum

Since no friction, total mechanical energy is conserved.

So:

- Position A:
- K > 0 (initial speed)
- U_g = 0
- U_s = 0 (no spring)
- E_tot = K

- Position B:
- Cart is still moving → K > 0
- Height increased → U_g > 0
- But since no friction, E_tot = K + U_g = constant
- So K decreases, U_g increases from A to B

- Position C:
- Not moving → K = 0
- Highest point → U_g is max
- Since E_tot conserved → U_g(C) = E_tot
- So U_g(C) = K(A)

Now let's fill in the bar charts:

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#### ✔ Completed Bar Charts for (a):

| Position | K | U_g | U_s | E_tot |
|---------|---------|---------|-----|--------|
| A | High | 0 | 0 | High |
| B | Medium | Medium | 0 | Same as A |
| C | 0 | High | 0 | Same as A |

> Note: Heights of bars should reflect conservation:
> - E_tot is constant across A, B, C
> - At A: all energy is kinetic
> - At C: all energy is potential (gravitational)
> - At B: mix of both

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b) No friction, initial height is zero

Description:
Cart starts at A (height = 0), goes up to B (top of hill), then down to C (same height as A). No friction.

Initial height is zero → U_g(A) = 0

Assume cart starts with some speed → K(A) > 0

No friction → mechanical energy conserved

#### Step-by-step:

- Position A:
- U_g = 0
- K > 0
- E_tot = K

- Position B:
- Highest point → U_g is maximum
- But cart is still moving → K > 0
- So K + U_g = E_tot (same as A)

- Position C:
- Same height as A → U_g = 0
- No friction → same speed as A → K = K(A)
- So E_tot = K

So:

- A: K high, U_g = 0
- B: K medium, U_g high (but K + U_g = constant)
- C: K high, U_g = 0

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#### ✔ Completed Bar Charts for (b):

| Position | K | U_g | U_s | E_tot |
|---------|---------|---------|-----|--------|
| A | High | 0 | 0 | High |
| B | Medium | High | 0 | Same |
| C | High | 0 | 0 | Same |

> Note: E_tot is constant. At B, U_g is highest, so K is lowest, but still positive. At C, back to original state.

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c) There is friction, car stopped at C, initial height is zero

Description:
Cart starts at A (height = 0), goes up to B, then to C (same height as A), but stopped at C due to friction.

Initial height = 0 → U_g(A) = 0
Friction present → mechanical energy is NOT conserved
Some energy is converted to thermal energy (not shown in bar charts)

At C, cart is stopped → K = 0
Height at C = height at A → U_g(C) = 0
So E_tot(C) = 0?

But wait: That would mean all energy was lost.

But the bar chart at A shows K > 0, U_g = 0 → E_tot > 0

So what happens?

- At A: K > 0, U_g = 0 → E_tot = K
- At B: Some K → U_g, but also loss due to friction → less total mechanical energy
- At C: Stopped → K = 0, U_g = 0 → E_tot = 0
→ So all mechanical energy was dissipated by friction

But is that possible?

Yes — if friction is strong enough, the cart may not make it back to same height or even stop before.

But here, it does reach C, which is at same height as A, and stops.

So:

- To go up to B, it needs energy
- Then to come down to C, but stopped at C → means kinetic energy at bottom is zero, so no speed

That implies: All initial mechanical energy was lost to friction

So:

- Position A:
- K > 0, U_g = 0 → E_tot = K

- Position B:
- Higher → U_g > 0
- Still moving → K > 0
- But due to friction, E_tot(B) < E_tot(A)
- So K + U_g < initial K

- Position C:
- U_g = 0
- K = 0
- So E_tot = 0

Thus, mechanical energy decreases due to friction.

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#### ✔ Completed Bar Charts for (c):

| Position | K | U_g | U_s | E_tot |
|---------|---------|---------|-----|--------|
| A | High | 0 | 0 | High |
| B | Medium | High | 0 | Less than A |
| C | 0 | 0 | 0 | 0 |

> Note:
> - E_tot decreases from A to B to C
> - At C, all mechanical energy is gone (dissipated as heat)
> - So E_tot(C) = 0

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Final Answer Summary:

#### (a) No friction, not moving at C

- A: K = full, U_g = 0, E_tot = full
- B: K = partial, U_g = partial, E_tot = full
- C: K = 0, U_g = full, E_tot = full

> All bars have same total height (conserved)

#### (b) No friction, initial height zero

- A: K = full, U_g = 0, E_tot = full
- B: K = partial, U_g = full, E_tot = full
- C: K = full, U_g = 0, E_tot = full

> E_tot constant; at B, U_g max, K min; at C, back to start

#### (c) With friction, stopped at C, initial height zero

- A: K = full, U_g = 0, E_tot = full
- B: K = partial, U_g = partial, E_tot = less than A
- C: K = 0, U_g = 0, E_tot = 0

> E_tot decreases due to friction; final mechanical energy = 0

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How to Draw the Bars:

For each position, draw bars such that:

- E_tot is the sum of K + U_g (since U_s = 0)
- E_tot is constant only when no friction
- E_tot decreases when friction is present

Use consistent scale: if K at A is 10 J, then total energy is 10 J.

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Final Sketch Guide (Text Representation):

#### (a)

```
Position A: K: [====] U_g: [] E_tot: [====]
Position B: K: [==] U_g: [==] E_tot: [====]
Position C: K: [] U_g: [====] E_tot: [====]
```

#### (b)

```
Position A: K: [====] U_g: [] E_tot: [====]
Position B: K: [==] U_g: [===] E_tot: [====]
Position C: K: [====] U_g: [] E_tot: [====]
```

#### (c)

```
Position A: K: [====] U_g: [] E_tot: [====]
Position B: K: [=] U_g: [==] E_tot: [==]
Position C: K: [] U_g: [] E_tot: []
```

> (E_tot decreasing, finally zero)

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✔ Answer Complete. You now know how to fill in each energy bar chart based on conservation of energy and presence of friction.