Question

haynes (lhh2749) - energy 1 - neff - (76523) 5
a. r₁

1. r₃
2. r₄

028 (part 1 of 2) 10.0 points
a bead slides without friction around a loop-the-loop. the bead is released from a height h from the bottom of the loop-the-loop which has a radius r.
which of the following diagrams best represents the kinetic energy of the bead versus time?
1.
2.
3.
4.
5.
6.
029 (part 2 of 2) 10.0 points
which of the following could represent the gravitational potential energy of the bead versus time?
1.
2.
3.
4.
5.
6.

Explanation:

Step1: Analyze Energy Conservation

The bead slides without friction, so mechanical energy (kinetic + gravitational potential) is conserved. Kinetic energy \( KE = \frac{1}{2}mv^2 \), gravitational potential energy \( PE = mgh_{\text{relative}} \).

Step2: Kinetic Energy vs Time (Part 028)

- Initially, the bead is at rest (released from height \( h \)), so initial \( KE = 0 \). As it falls, \( KE \) increases (speed increases) until it reaches the bottom of the loop. Then, as it moves up the loop, \( KE \) decreases (speed decreases) because \( PE \) increases (height increases). At the top of the loop, \( KE \) is minimum (but not zero, since it must have enough speed to stay on the loop). Then, as it moves down the loop, \( KE \) increases again, and after the loop, as it moves along the flat part, \( KE \) should remain constant (since height is constant, no change in \( PE \), so \( KE \) is constant).
- Looking at the options: Option 4 (the fourth graph for part 028) starts at 0, increases, decreases, increases, then levels off (constant), which matches this behavior. Wait, no—wait, initial \( KE \) is 0? Wait, the bead is released from rest? The problem says "released from a height \( h \)", so initial velocity is 0, so initial \( KE = 0 \). So the graph should start at 0. Then, as it falls, \( KE \) increases (speed up), then up the loop: \( KE \) decreases (speed down), then down the loop: \( KE \) increases (speed up), then after the loop, moving on the flat, \( KE \) is constant. So the graph that starts at 0, has a peak, then a trough, then a peak, then constant? Wait, the options for part 028: let's re-examine. Option 4: starts at 0, has a wave (increase, decrease, increase) then maybe constant? Wait, maybe I misread. Wait, the first option (1) has \( KE \) starting high, which is wrong (initial \( KE = 0 \)). Option 2: starts at 0, has a wave (increase, decrease, increase) but does it level off? Option 4: starts at 0, wave, then maybe constant? Wait, the correct graph for \( KE \) vs time: initial \( KE = 0 \), so starts at 0. Then, as it falls, \( KE \) increases (since speed increases), then when going up the loop, \( KE \) decreases (speed decreases), then when going down the loop, \( KE \) increases (speed increases), then after the loop, on the flat part, \( KE \) is constant (since height is constant, no change in \( PE \), so \( KE \) doesn't change). So the graph should start at 0, have a peak (bottom of loop), trough (top of loop), peak (bottom of loop exit), then constant. Looking at the options for part 028, option 4 (the fourth graph) starts at 0, has a wave (up, down, up) then maybe constant? Or option 2? Wait, maybe I made a mistake. Wait, the problem is part 028: "Which of the following diagrams best represents the kinetic energy of the bead versus time?" The options:

1. Graph starts high, then waves, then zero? No, initial \( KE = 0 \), so eliminate.
2. Graph starts at 0, waves (up, down, up) but no constant? No, after the loop, it should be constant.
3. Graph starts high, waves, then constant? No, initial \( KE = 0 \).
4. Graph starts at 0, waves (up, down, up) then constant? Maybe.
5. Graph starts at 0, peak, then zero? No.

Wait, maybe the correct answer for part 028 is option 4? Wait, no, maybe I messed up. Alternatively, for part 029 (gravitational potential energy vs time):

Gravitational potential energy \( PE = mgh(t) \), where \( h(t) \) is the height above the bottom. Initially, \( h = h \), so \( PE \) is maximum. As the bead falls, \( h \) decreases, so \( PE \) decreases. Then, as it moves up the loop, \( h \) increases, so \( PE \) increases. At the top of the loop, \( h \) is maximum (2r above bottom), so \( PE \) is maximum there. Then, as it moves down the loop, \( h \) decreases, so \( PE \) decreases. After the loop, moving on the flat, \( h \) is constant (equal to the bottom height), so \( PE \) is constant (minimum, since \( h = 0 \) relative to bottom? Wait, the bottom of the loop is the reference, so after the loop, the height is constant (flat), so \( PE \) is constant (and lower than initial). So the \( PE \) vs time graph should start high, decrease, increase (to a peak at the top of the loop), decrease, then level off (constant, low). Looking at the options for part 029:

1. Graph starts high, has a dip, then a peak, then constant? No.
2. Starts at 0, increases: wrong (initial \( PE \) is high).
3. Starts high, decreases, increases (peak), decreases, then constant: this matches. So option 3 for part 029.

But the question here—wait, the user's problem: maybe we need to solve part 028 first. Wait, the user's image has two parts: 028 (kinetic energy vs time) and 029 (gravitational potential energy vs time). Let's focus on part 028 first.

Step3: Re-evaluate Part 028

- Initial \( KE = 0 \) (released from rest), so the graph must start at \( KE = 0 \) (y=0 at t=0).
- As the bead falls (down the first slope), speed increases, so \( KE \) increases (graph rises).
- As it moves up the loop (toward the top), speed decreases (since height increases, \( PE \) increases, so \( KE \) decreases: graph falls).
- At the top of the loop, \( KE \) is minimum (but not zero, since it needs centripetal force to stay on the loop: \( v \geq \sqrt{gr} \)).
- As it moves down the loop (from top to bottom of the loop), speed increases, so \( KE \) increases: graph rises.
- After the loop, it moves on a flat surface (constant height), so \( PE \) is constant, hence \( KE \) is constant (graph becomes horizontal, constant).

Looking at the options for part 028:
- Option 1: Starts at high \( KE \) (wrong, initial \( KE = 0 \)).
- Option 2: Starts at 0, has a wave (rise, fall, rise) but does it level off? No, it keeps waving.
- Option 3: Starts at 0, rises to a peak, then falls, then rises, then levels off? Wait, no, option 3's graph looks like a single peak then a drop? No, the options: let's see the numbers. The fifth option (5) for part 028: starts at 0, has a peak, then a trough, then a peak, then maybe constant? Wait, maybe the correct answer for part 028 is option 4? Wait, no—wait, the fourth option (4) for part 028: the graph starts at 0, has a wave (increase, decrease, increase) and then maybe constant? Or option 4 is the one that starts at 0, increases, decreases, increases, then levels off. Yes, that matches: initial \( KE = 0 \), then up (falling), down (climbing loop), up (descending loop), then constant (flat part). So option 4 for part 028.

For part 029 (gravitational potential energy vs time):
- Initial \( PE \) is high (at height \( h \)).
- As it falls, \( PE \) decreases (height decreases), so graph falls.
- As it climbs the loop, \( PE \) increases (height increases), so graph rises (to a peak at the top of the loop).
- As it descends the loop, \( PE \) decreases (height decreases), so graph falls.
- After the loop, on the flat, height is constant, so \( PE \) is constant (graph levels off, low).

Looking at the options for part 029:
- Option 1: Starts high, dips, peaks, then constant? No, the peak is too low.
- Option 2: Starts at 0, increases: wrong (initial \( PE \) is high).
- Option 3: Starts high, decreases, increases (peak at top of loop), decreases, then levels off (constant, low): this matches. So option 3 for part 029.

But the user's question—wait, the problem is to solve these? Let's confirm.

Answer (for part 028): 4. (the fourth graph for part 028)

Answer (for part 029): 3. (the third graph for part 029)

Wait, but the user's image: for part 028, the options are numbered 1-5. Let's check the graphs:

Part 028 options:
1. Graph starts at high KE, waves, then zero.
2. Graph starts at 0, waves (up, down, up) without leveling.
3. Graph starts at 0, single peak, then drop.
4. Graph starts at 0, waves (up, down, up) then levels off.
5. Graph starts at 0, small peak, then trough, then peak, then zero.

So the correct one for KE vs time is option 4 (starts at 0, increases, decreases, increases, then constant), because after the loop, KE is constant (flat part).

For part 029 (PE vs time):
Options:
1. Starts high, dips, peaks, then constant (but peak is low).
2. Starts at 0, increases: wrong.
3. Starts high, decreases, increases (peak at top of loop), decreases, then constant (low): correct.
4. Starts high, decreases, increases, decreases: wrong (should level off).
5. Linear decrease: wrong.

So part 028 answer: 4. (the fourth graph)
Part 029 answer: 3. (the third graph)

Answer:

(for part 028): 4. (the fourth graph for part 028)