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(x), 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 x₂ 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 decreasing and approaching a horizontal line
2. none of these graphs is correct.
3. graph of u vs t with a curve increasing and approaching a horizontal line
4. graph of u vs t with a straight line increasing
5. graph of u vs t with a straight line decreasing
6. graph of u vs t with a curve approaching a horizontal line from below
7. graph of u vs t with a curve increasing from a low value
021 (part 1 of 2) 10.0 points
the following graph represents a hypothetical

Question

Accepted Answer

### 017 (part 1 of 2) Solution:
# Explanation:
## Step1: Recall Work Formula
Work $ W $ is force $ F $ times distance $ d $, $ W = F \cdot d $.
## Step2: Substitute Values
Given $ F = 150 \, \text{N} $, $ d = 5.00 \, \text{m} $. So $ W = 150 \times 5.00 $.
# Answer:
$ 750 \, \text{J} $

### 017 (part 2 of 2) Solution:
# Explanation:
## Step1: Constant Velocity Implies Equilibrium
Net force is zero, so applied force $ F $ equals friction force $ f $, $ f = F = 150 \, \text{N} $. Friction $ f = \mu_k N $, and normal force $ N = mg $ (since horizontal surface, $ N $ balances weight).
## Step2: Calculate Normal Force
$ m = 50.0 \, \text{kg} $, $ g = 9.81 \, \text{m/s}^2 $, so $ N = 50.0 \times 9.81 = 490.5 \, \text{N} $.
## Step3: Solve for $ \mu_k $
From $ f = \mu_k N $, $ \mu_k = \frac{f}{N} = \frac{150}{490.5} \approx 0.306 $.
# Answer:
$ \approx 0.306 $

### 020 Solution:
# Brief Explanations:
Gravitational potential energy (GPE) between two masses: $ U = -\frac{G M m}{r} $, where $ r $ is distance. As the small mass falls (moves towards the large mass), $ r $ decreases, so $ |U| $ increases (becomes less negative). At $ t = 0 $, $ r $ is very large, so $ U \approx 0 $ (since $ 1/r \approx 0 $). As time increases, $ r $ decreases, $ U $ becomes more negative? Wait, no—wait, when released from rest far away, it accelerates towards the large mass, so $ r $ decreases. The GPE formula is $ U = -\frac{G M m}{r} $, so as $ r $ decreases (time increases, moving closer), $ U $ becomes more negative (decreases? Wait, no: at $ r \to \infty $, $ U = 0 $; as $ r $ decreases, $ U $ becomes negative and its magnitude increases. But the graphs: let's check. Wait, the options: graph 5? Wait, no—wait, the small mass is moving towards the large mass, so $ r(t) $ is a function that decreases over time (approaching the large mass). So $ U(t) = -\frac{G M m}{r(t)} $. As $ t $ increases, $ r(t) $ decreases, so $ U(t) $ decreases (becomes more negative). So the graph should show $ U $ starting near 0 (at $ t=0 $, $ r $ large) and decreasing over time. Looking at the options, graph 5 has $ U $ starting at a dashed line (maybe 0) and decreasing. Wait, but let's re-express: the formula is $ U = -\frac{G M m}{r} $, so as $ r $ decreases (time increases, moving closer), $ U $ becomes more negative (decreases). So the correct graph is the one where $ U $ starts near 0 and decreases, which is graph 5? Wait, no—wait the options:

1. $ U $ decreases then levels? No, because $ r $ keeps decreasing (approaching the large mass, but actually, when it reaches the large mass, but in reality, it accelerates, so $ r(t) $ is a function that decreases with time, but the GPE as a function of time: let's think about the velocity. The small mass accelerates, so $ dr/dt $ (magnitude) increases (since force increases as $ r $ decreases, so acceleration increases, speed increases, so $ r $ decreases faster over time). So $ U(t) = -\frac{G M m}{r(t)} $, and since $ r(t) $ is decreasing at an increasing rate (because speed increases), the rate of change of $ U $ (slope) should increase in magnitude (since $ U \propto -1/r $, so $ dU/dt = \frac{G M m}{r^2} \cdot dr/dt $, and $ dr/dt $ (magnitude) increases, $ r $ decreases, so $ dU/dt $ (magnitude) increases, meaning the graph of $ U $ vs $ t $ should be a curve that becomes steeper (more negative) as $ t $ increases. But looking at the options, graph 5 is a straight line? No, wait the options:

Wait the options:

1. $ U $ starts high, decreases, then levels? No.

3. $ U $ increases (becomes less negative) over time? No, because $ r $ is decreasing, so $ U $ should become more negative.

5. $ U $ starts at a dashed line (maybe 0) and decreases, but is it a straight line? Wait, no—wait the correct analysis: the gravitational potential energy between two masses is $ U = -\frac{G M m}{r} $. When the small mass is released from rest at $ r \to \infty $, $ U \approx 0 $. As it moves towards the large mass, $ r $ decreases, so $ U $ becomes negative (and its magnitude increases). The velocity of the small mass increases as it falls (since gravitational force does work on it), so $ r(t) $ decreases at an increasing rate (because acceleration increases as $ r $ decreases, so speed increases, so $ r $ gets smaller faster over time). Therefore, $ U(t) = -\frac{G M m}{r(t)} $, and since $ r(t) $ is a function that decreases with time, and the rate of decrease of $ r(t) $ increases (speed increases), the rate of change of $ U(t) $ (the slope of $ U $ vs $ t $) should increase in magnitude (because $ dU/dt = \frac{G M m}{r^2} \cdot |dr/dt| $, and $ |dr/dt| $ increases, $ r $ decreases, so $ dU/dt $ increases in magnitude). So the graph should be a curve that starts near 0 and becomes steeper (more negative) as $ t $ increases. But looking at the options, none of the graphs (except maybe graph 5? Wait no—wait the options:

Wait the options are:

1. $ U $ decreases then levels (asymptote? No, because it should keep decreasing as $ r $ approaches the large mass, but actually, when it hits the large mass, but in the problem, it's a "very great distance" initially, so as $ t $ increases, $ r $ decreases, $ U $ becomes more negative. But the graphs:

Graph 5: $ U $ starts at a dashed line (maybe 0) and decreases linearly? No, it should be a curve. Wait, maybe the correct answer is "None of these graphs is correct" (option 2)? Wait no—wait let's re-express. Wait the formula is $ U = -\frac{G M m}{r} $, so $ U(t) = -\frac{G M m}{r(t)} $. The small mass is accelerating, so $ r(t) $ is a function where $ dr/dt = -v(t) $, and $ v(t) $ increases (since acceleration $ a = F/m = (G M m / r^2)/m = G M / r^2 $, which increases as $ r $ decreases). So $ r(t) $ is a function that decreases with time, and the rate of decrease (speed) increases. Therefore, $ U(t) $ is a function that starts at 0 (when $ t=0 $, $ r \to \infty $) and becomes more negative over time, with the slope (rate of change of $ U $) becoming steeper (more negative) as $ t $ increases. Now, looking at the graphs:

- Graph 1: $ U $ decreases then levels (asymptote to a dashed line). No, because $ r $ keeps decreasing, so $ U $ should keep decreasing (more negative).

- Graph 3: $ U $ increases (becomes less negative) over time. No, that's the opposite.

- Graph 4: $ U $ increases linearly from 0. No.

- Graph 5: $ U $ decreases linearly from 0. But we need a curve (since $ v(t) $ increases, so $ r(t) $ decreases faster, so $ U(t) $ should decrease faster, i.e., the slope should increase in magnitude). So a linear graph (constant slope) is incorrect.

- Graph 6: $ U $ has a U-shape? No.

- Graph 7: $ U $ increases from a low value. No.

So none of the graphs correctly represent $ U(t) $, because the correct graph should be a curve with increasing slope (more negative) over time, but the options don't have that. So the answer is option 2: None of these graphs is correct.

# Answer:
2. None of these graphs is correct.

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QUESTION IMAGE

Question

Explanation:

017 (part 1 of 2) Solution:

Step1: Recall Work Formula

Step2: Substitute Values

Step1: Constant Velocity Implies Equilibrium

Step2: Calculate Normal Force

Step3: Solve for \( \mu_k \)

Answer:

017 (part 2 of 2) Solution: