haynes (hlh2749) - energy 1 - neff - (76523)
answer in units of j.
016 (part 3 of 3) 10.0 points
c) find the coefficient of kinetic friction be-
tween the flight bag and the floor.
017 (part 1 of 2) 10.0 points
a horizontal force of 150 n is used to push a
50.0 kg packing crate a distance of 5.00 m on
a rough horizontal surface.
the acceleration of gravity is 9.81 m/s².
if the crate moves with constant velocity,
calculate
a) the work done by the force.
answer in units of j.
018 (part 2 of 2) 10.0 points
b) the coefficient of kinetic friction.
019 10.0 points
a conservative force has the potential energy
function u(r), shown by the graph. a particle
moving in one dimension under the influence
of this force has kinetic energy 1.0 joule when
it is at position x₁
potential energy vs position
graph of potential energy (j) vs position with x₀, x₁, x₂, x₃ on x - axis and potential energy from -1 to 1 on y - axis
which of the following is a correct state-
ment about the motion of the particle?
1. it moves to the right of x₃ and does not
return.
2. it comes to rest at either x₀ or x₃ and
remains at rest.
3. it moves to the left of x₀ and does not
return.
4. it cannot reach either x₀ or x₂.
5. it oscillates with maximum position x₂
and minimum position x₀.
020 10.0 points
a small mass is released from rest at a very
great distance from a much larger stationary
mass.
which of the following graphs best repre-
sents the gravitational potential energy u of
the system of masses as a function of t?
1. graph of u vs t with a curve starting high and approaching a horizontal dashed line
2. none of these graphs is correct.
3. graph of u vs t with a curve starting low and approaching a horizontal dashed line from below
4. graph of u vs t with a straight line starting at origin and increasing
5. graph of u vs t with a straight line starting high and decreasing
6. graph of u vs t with a curve starting high and curving down
7. graph of u vs t with a curve starting low and curving up
021 (part 1 of 2) 10.0 points
the following graph represents a hypothetical

Question

Accepted Answer

### 017 (part 1 of 2) Solution:
# Answer:
750 J
# Explanation:
## Step1: Recall work formula
Work $ W = F \cdot d \cdot \cos\theta $, where $ F $ is force, $ d $ is distance, $ \theta $ is angle between force and displacement.
## Step2: Identify values
Here, $ F = 150 \, \text{N} $, $ d = 5.00 \, \text{m} $, $ \theta = 0^\circ $ (horizontal force, horizontal displacement), so $ \cos 0^\circ = 1 $.
## Step3: Calculate work
$ W = 150 \, \text{N} \times 5.00 \, \text{m} \times 1 = 750 \, \text{J} $.

### 018 (part 2 of 2) Solution:
# Answer:
0.306
# Explanation:
## Step1: Analyze forces (constant velocity)
Net force $ F_{\text{net}} = 0 $, so applied force $ F $ equals kinetic friction force $ f_k $. Kinetic friction $ f_k = \mu_k \cdot N $, where $ N $ is normal force. On horizontal surface, $ N = mg $.
## Step2: Identify values
$ m = 50.0 \, \text{kg} $, $ g = 9.81 \, \text{m/s}^2 $, $ F = 150 \, \text{N} $. So $ N = 50.0 \times 9.81 = 490.5 \, \text{N} $, and $ f_k = F = 150 \, \text{N} $.
## Step3: Solve for $ \mu_k $
$ \mu_k = \frac{f_k}{N} = \frac{150}{490.5} \approx 0.306 $.

### 019 Solution (Multiple - Choice Analysis):
# Answer:
5. It oscillates with maximum position $ x_2 $ and minimum position $ x_0 $.
# Brief Explanations:
- Total mechanical energy $ E = K + U $. At $ x_1 $, $ K = 1 \, \text{J} $, so $ E = 1 + U(x_1) $. From the potential energy graph, $ U(x_1) \approx - 1 \, \text{J} $, so $ E \approx 0 \, \text{J} $.
- The particle moves where $ U(x) \leq E $. The potential energy curve has a minimum between $ x_0 $ and $ x_2 $, and a maximum (or plateau) after $ x_2 $. With $ E \approx 0 $, the particle is bound between $ x_0 $ (where $ U $ is high) and $ x_2 $ (where $ U $ rises to near $ E $), so it oscillates between $ x_0 $ (min position) and $ x_2 $ (max position).

### 020 Solution (Multiple - Choice Analysis):
# Answer:
7. (The graph with $ U $ approaching a constant from below as $ t $ increases)
# Brief Explanations:
- Gravitational potential energy between two masses is $ U = -\frac{G M m}{r} $, where $ r $ is distance between them. The small mass is released from rest at large $ r $ (so initial $ U \approx 0 $, since $ r \to \infty $, $ U \to 0 $) and moves toward the large mass ( $ r $ decreases). As $ r $ decreases, $ |U| $ increases (becomes more negative initially? No—wait, $ U $ is negative, and as $ r $ decreases, $ U $ becomes more negative (smaller) in magnitude? Wait, no: $ U = -\frac{G M m}{r} $, so when $ r $ is large, $ U \approx 0 $; as $ r $ decreases (mass moves closer), $ U $ becomes negative (since $ G, M, m $ positive). But the mass accelerates, so $ r(t) $ decreases non - linearly (since acceleration depends on $ r^2 $). The potential energy $ U(t) $ will approach $ -\frac{G M m}{r_{\text{final}}} $ (but since the large mass is stationary and much larger, the small mass will have $ r $ approaching a value? Wait, no—actually, the gravitational potential energy as a function of time: the mass starts at rest far away ( $ U \approx 0 $ ), moves towards the large mass, so $ r $ decreases, $ U = -\frac{G M m}{r} $ becomes more negative (smaller) in value, but the rate of change of $ r $ (velocity) increases (since acceleration $ a=\frac{G M}{r^2} $ increases as $ r $ decreases). Wait, maybe the graph: as time increases, $ r $ decreases, so $ U = -\frac{G M m}{r} $ approaches a constant (since as the small mass gets closer, $ r $ changes less? No, actually, the correct graph should show $ U $ (which is negative) approaching a constant (the potential energy at close distance) as $ t $ increases, with the curve having a shape where $ U $ starts near 0 and becomes more negative (but the option 7 is a graph where $ U $ is below a dashed line and approaches it, which matches: initial $ U \approx 0 $ (dashed line maybe at $ U = 0 $? No, wait, maybe the dashed line is the potential energy at infinity (0), and as the mass moves, $ U $ becomes negative, but the graph 7 has $ U $ approaching a constant from below, which is consistent with $ U = -\frac{G M m}{r} $ as $ r $ decreases ( $ U $ becomes more negative, but the rate of change slows as $ r $ approaches the large mass? No, actually, the correct graph is the one where $ U $ starts near 0 (at large $ r $) and decreases (becomes more negative) but the curve's slope (rate of change of $ U $) increases initially (since velocity increases) then maybe? Wait, the key is: the gravitational potential energy as a function of time for a mass falling into a gravitational field (from rest at infinity) will have $ U(t) $ approaching $ -\frac{G M m}{R} $ (where $ R $ is the radius of the large mass, if it's a planet) as $ t $ increases, and the graph that shows $ U $ starting near 0 and approaching a constant (more negative) with the curve's shape matching the non - linear motion (acceleration increasing, so $ r(t) $ is a curve, $ U(t) $ is a curve approaching a constant). The 7th graph (assuming the last one with $ U $ curving towards a constant) is correct because as the mass moves closer, $ r $ decreases, $ U $ becomes more negative (but the graph's y - axis for $ U $: if the dashed line is the potential at infinity (0), then $ U $ starts near 0 and goes to a constant (the potential at the surface, say), and the graph 7 shows $ U $ approaching a constant from below (since $ U $ is negative) as $ t $ increases, which matches the behavior.

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

017 (part 1 of 2) Solution:

Step1: Recall work formula

Step2: Identify values

Step3: Calculate work

018 (part 2 of 2) Solution:

Step1: Analyze forces (constant velocity)

Step2: Identify values

Step3: Solve for \( \mu_k \)

019 Solution (Multiple - Choice Analysis):

020 Solution (Multiple - Choice Analysis):

Answer: