Read each description below. Choose the force diagram (free-body diagram) that best represents the description. You may neglect the effects of air resistance. |
A. Mark's physics book is on the desk. Mark pushes horizontally (from the side) on the book, but the book does not move. |
B. Kendall's physics book is on the desk. Kendall pushes horizontally (from the side) on the book, and the book moves across the desk at a constant speed. |
C. Jenny pushes a cart at constant velocity down the grocery store aisle. |
D. Eddy is driving his Camaro down a straight, level road while on cruise control. |
E. Ryan is accelerating his Dakota down a straight, level road. |
F. Jackie's Viper is parked on the shoulder of the road while she changes a flat tire. |
Answers
Answer:
A. Mark's physics book is on the desk. Mark pushes horizontally (from the side) on the book, but the book does not move.
DIAGRAM B
B. Kendall's physics book is on the desk. Kendall pushes horizontally (from the side) on the book, and the book moves across the desk at a constant speed. |
DIAGRAM B
C. Jenny pushes a cart at constant velocity down the grocery store aisle.
DIAGRAM B
D. Eddy is driving his Camaro down a straight, level road while on cruise control.
DIAGRAM B
E. Ryan is accelerating his Dakota down a straight, level road.
DIAGRAM C
F. Jackie's Viper is parked on the shoulder of the road while she changes a flat tire.
DIAGRAM A
Explanation:
A. Since the book does not move even after being pushed it is leveled in all forces.
B. Since it is pushed at a constant speed, all forces are equal.
C. Since it is pushed at a constant speed, all forces are equal.
D. Since it is driven at a constant speed on cruise control, all forces are equal.
E. Since he is accelerating, which means it is being driven much faster than the friction that is pushing the car back, the forces are unequal.
F. The only forces acting on the is the car on the ground and the ground on the car, there is no other forces of friction pushing it back or the car being pushed forward. It is parked.
In option A, the book does not move, so the force diagram would show the normal force and the force of friction. In option B, the book moves at a constant speed, so the force diagram would show the normal force and the force of friction in the same direction as the applied force. Option C involves a cart moving at constant velocity, which would have a force diagram showing the forward push and the force of friction. Options D and E involve cars on level roads, with D having a balanced force diagram and E having unbalanced forces causing acceleration. Option F involves a parked car with a balanced force diagram.
Explanation:The force diagram that best represents the scenario described in option A, where Mark's physics book is on the desk and Mark pushes horizontally on the book but it does not move, would be a free-body diagram showing only the normal force and the force of friction acting on the book. Since the book does not move, the forces are balanced, with the force of friction equal in magnitude and opposite in direction to the horizontal force applied by Mark.
In option B, where Kendall's physics book is on the desk and Kendall pushes horizontally on the book and it moves across the desk at a constant speed, the force diagram would also include the normal force and the force of friction, but the force of friction would be in the same direction as Kendall's applied force, since the book is moving. The forces are balanced and the net force is zero.
In option C, where Jenny pushes a cart at constant velocity down the grocery store aisle, the force diagram would include a force in the forward direction representing Jenny's push, as well as the force of friction opposing the motion. The forces are balanced and the net force is zero.
In option D, where Eddy is driving his Camaro down a straight, level road while on cruise control, there would be a force diagram showing the force of the engine pushing the car forward, the force of friction opposing the motion, and the normal force. The forces are balanced and the net force is zero.
In option E, where Ryan is accelerating his Dakota down a straight, level road, the force diagram would include the force of the engine pushing the car forward, the force of friction opposing the motion, and the normal force. The net force is non-zero, as the forces are unbalanced and causing the acceleration.
In option F, where Jackie's Viper is parked on the shoulder of the road while she changes a flat tire, there would be a force diagram showing only the force of gravity acting on the car, the normal force from the ground, and the force of friction from the ground. The forces are balanced and the net force is zero.
Related Questions
all waves carry a) energy B) light C) matter D) particles
Answers
All waves, regardless of their type, carry energy from one place to another without transporting matter. They do not all carry light, matter, or particles.
Explanation:All waves, regardless of whether they are light waves, sound waves, electromagnetic waves, or even water waves, carry energy. Therefore, the answer to your question is a) energy.
Waves transfer energy from one place to another without transporting matter. They do this through oscillations and disturbances in a medium. For example, when you throw a stone into a pond, the waves created by the stone do not transport the water from one side of the pond to another. Instead, they transport the energy created by the stone's impact.
Not all waves carry light, matter, or particles. Light is specifically carried by electromagnetic waves, while particles and matter are not typically associated with waves in the conventional sense.
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Related Rates: the hypotenuse of an isosceles right triangle is increasing at a rate of 2mm/s. at what rate is the area of the triangle increasing when the length of one of the equal sides is 10mm? Related Rates: the hypotenuse of an isosceles right triangle is increasing at a rate of 2mm/s. at what rate is the area of the triangle increasing when the length of one of the equal sides is 10mm?
Answers
Below is the solution:
Call the length of the legs x and the hypotenuse z, then we have z2=2x2(Pythagorean Theorem)
We know that the rate of change of x with respect to time, t, is dx/dt=2 mm/s.
We want to find dz/dt at the instant when x=10 mm
Differentiate z2=2x2 implicitly with respect to time, t:
d/dt(z2) = d/dt(2x2)
2z dz/dt = 4x dx/dt
z dz/dt = 2x dx/dt
z=10√2 mm
10√2 mm)dz/dt = 2(10 mm)(2 mm/s)
dz/dt = 2√2 mm/s
Final answer:
The rate at which the area of the triangle increases when the length of one of the equal sides is 10mm is 0 mm²/s.
Explanation:
To find the rate at which the area of the triangle is increasing, we can use the formula for the area of a right triangle, which is (1/2) * base * height. Since the triangle is isosceles, the base and height are equal. Let's call the length of the equal side x. The hypotenuse is also related to x by the Pythagorean theorem, which states that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
We can set up the equations:
x² + x² = c²
2x² = c²
Taking the derivative of both sides concerning time, we get:
4x * (dx/dt) = 2c * (dc/dt)
Plugging in the given values, where dx/dt = 0 (since x is constant) and dc/dt = 2mm/s, we can solve for the rate at which the area is increasing:
4(10) * (0) = 2(10.3) * (dA/dt)
0 = 20.6 * (dA/dt)
(dA/dt) = 0 mm²/s
A star that's actively fusing hydrogen into helium is aa. supergiant.
b. main-sequence star.
c. red giant.
d. white dwarf.
Answers
It is the stage, which comes in life cycle of every star, when they perform Nuclear fusion inside their core.
Hope this helps!
actively fumes hydrogen to helium
What is series circuit
Answers
Nuclear power plants in use today rely on ________.
Answers
A garage door opener has a power rating of 350 watts.
If the door is in operation for 30 seconds, how many joules of energy are used? (Hint: Remember that 1 watt = 1 joule/sec) 350 J 11.7 J 10,500 J 0.086 J
Answers
Substitute the known value. which is P = 350 watt, & T = 30 sec
350 = E/30
E = 350 * 30
E = 1050 Joules.
So, your answer is 1050 Joules.
Hope this helps! Keep studying!
Which best describes a reference frame?
A. a distance over which an object moves
B. the difference between start and end velocity
C. a position from which something is observed
D. the average speed of a group of objects
Answers
By definition we have to:
A system or frame of reference are those conventions used by an observer (usually standing at a point on the ground) to be able to measure the position and other physical magnitudes such as speed and acceleration of one or several objects.
The numerical value of some magnitudes can also be relative to the reference system when we refer to the relative movement. There are always mathematical relationships between the observer and the relative magnitudes.
Answer:
C. a position from which something is observed
You want to use a lever to lift a 2000 N rock. The maximum force you can exert is 500 N. Draw a lever to lift the rock. Label the input force, output force, fulcrum, input arm, and output arm. Specify measurements for the input and output arms. State the mechanical advantage of your lever.
Answers
Explanation of how a lever works and how to calculate its mechanical advantage when lifting a load with a given input force.
A lever is a simple machine that consists of a rigid bar pivoted at a fixed point called the fulcrum. It helps in exerting a smaller force over a longer distance to lift a heavier load over a shorter distance.
The mechanical advantage of a lever is calculated by dividing the length of the effort arm by the length of the load arm (MA = Le/Lr). It tells how many times a lever multiplies the input force to lift a load.
In the described scenario, with an input force of 500 N and a load of 2000 N, the mechanical advantage of the lever can be calculated using the formula MA = Load / Effort, which will be 4.
The number of times that a wave vibrates in a unit of time is called
A. amplitude
B. frequency
C. wavelength
D. interference ...?
Answers
Casey has an open book test in his physics class tomorrow and will need to use many formulas. Which of the following is the best strategy Casey could use to organize his formulas?
Answers
Casey can organize physics formulas by categorizing them based on topics, arranging them sequentially, using color-coded tabs, prioritizing key formulas, and creating a concise index for quick reference during the open book test.
Casey can adopt a systematic approach to organize his physics formulas for the upcoming open book test, promoting efficiency and accessibility during the examination. Firstly, he can categorize the formulas based on the topics or chapters they belong to. This helps create a structured framework that aligns with the course syllabus, making it easier for Casey to locate specific formulas relevant to different sections of the test.
Within each category, Casey may further arrange the formulas in a logical sequence, following the order of topics covered in the class. This sequential arrangement aids in maintaining a smooth flow while navigating through the formulas during the test, reducing the time spent searching for the right information. Additionally, he could use color-coded tabs or highlighters to distinguish between different formula categories, providing a visual aid for quick identification.
Considering the importance of some formulas over others, Casey might prioritize key formulas or those frequently used in class. By placing these prominently at the beginning of each category, he ensures immediate access to crucial information. Moreover, creating a concise index or summary at the front of his formula compilation can serve as a quick reference guide, aiding Casey in quickly identifying the page numbers corresponding to specific formulas.
By implementing these strategies, Casey can enhance his organization of physics formulas, facilitating a more efficient and stress-free experience during the open book test.
The question probable maybe: What organizational method could Casey employ to effectively categorize and arrange the various physics formulas for his open book test, ensuring easy access and streamlined use during the examination?
What type of energy is held in the steam produced in a power plant?
A. nuclear
B. heat
C. light What type of energy is held in the steam produced in a power plant?
A. nuclear
B. heat
C. light
Answers
Heat is the type of energy that is held in the steam produced in a power plant.
I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
Answer: B. heat
Explanation:
The nuclear reaction involves the simultaneous fission and fusion of the heavy metal nuclei such as uranium in control conditions of the nuclear reactor so as to produce energy in the form of heat. The heat or thermal energy is used to boil the water and convert it the form of steam. The steam is used to run the turbine of the generator which produces electricity.
a 0.15 kg meter stick is balanced with the pivot point at the 18 cm mark. A weight of 3.2N is hung from the shorter end. Where should the weight be positioned in order to balance the meter stick?
...?
Answers
(0.15 x g)/100 = 0.0147N. per cm. length.
Torque short end = (0.0147 x 9) = 0.1323N/cm.
Torque long end = ((100 - 18)/2) x 0.0147 = 0.6027N/cm.
Difference = (0.6027 - 0.1323) = 0.4704N/cm.
(0.4704/3.2) = 0.147cm. from the 18cm. mark, = 17.853cm. from the 0 end of the stick.
Answer:
At 15cm to the left of the pivot
Explanation:
The weight 0.15kg(1.5N) and 3.2N are balanced on a meter stick. This weights are parallel to each other on the stick. Questions on parallel forces acting on a body are solved using the *principle of moment."
Moment is the turning effect of force about a point.
Principle of moment states that the sum of clockwise moment is equal to the sum of anticlockwise moments.
According to the question, the meter (100cm) stick is balanced with a pivot @ 18cm mark {this is the shorter end}
If the 3.2N weight is hung at xcm from the shorter end, this weight will turn about the pivot in the anticlockwise direction.
Since moment = Force × perpendicular distance from the force
Moment of the 3.2N weight = 3.2N × x = 3.2x
The weight of the stick (1.5N) will be positioned at the center of the stick i.e @ the 50cm mark which is at 32cm to the right of the pivot.
The weight of the stick will turn in the clock wise direction
The moment of the weight in the clockwise direction = 1.5×(50-18) = 1.5×32 = 48Ncm
Using the principle of moment, we will equate both moments to have;
3.2x = 48
x = 48/3.2
x = 15cm
This means the 3.2N weight will be positioned at the 15cm mark to the left if the pivot to balance the meter stick.
Present at least one example that illustrates acceleration. Is this a scalar or vector quantity? Explain why. ...?
Answers
Example :
a Car stopping at the rate of 5 m/s. In this example, the direction of acceleration is the opposite of the direction of velocity, and the magnitude is 5 m/s
Hope this helps
Answer:
Vector
Explanation:
Acceleration is defined as the rate of change of velocity. As velocity is a vector quantity, so acceleration is also a vector quantity.
If a car starts from rest and attains some velocity after some time, then car is accelerating and the firection of acceleration is same as the direction of velocity.
If a car is moving and after applying the brakes it come to rest it means the motion of car has negative acceleration which means the direction of velocity and the direction of acceleration is opposite to each other.
. On level ground a shell is fired with an initial velocity of
at 60.0° above the horizontal and feels no appreciable air
resistance. (a) Find the horizontal and vertical components of the
shell’s initial velocity. (b) How long does it take the shell to reach
its highest point? (c) Find its maximum height above the ground.
(d) How far from its firing point does the shell land? (e) At its
highest point, find the horizontal and vertical components of its
acceleration and velocity.
Answers
Using basic kinematics and trigonometry in projectile motion, we find the horizontal and vertical velocity components, time to reach the highest point, and the maximum height. We then determine the range of the projectile and the acceleration and velocity components at the highest point.
Let's solve the question step by step:
(a) Horizontal and Vertical Components of Velocity
To find the components of the shell's initial velocity, we can use trigonometry:
• Horizontal component (vx): vx = v * cos(θ) = 70.0 m/s * cos(60.0°)
• Vertical component (vy): vy = v * sin(θ) = 70.0 m/s * sin(60.0°)
(b) Time to Reach the Highest Point
The time to reach the highest point can be determined using the formula: t = vy / g, where g is the acceleration due to gravity (9.8 m/s2).
(c) Maximum Height
The maximum height is given by H = (vy2) / (2g).
(d) Horizontal Range
The horizontal range can be calculated once we have the total time of flight and the horizontal velocity: range = vx * total time of flight.
(e) Components of Acceleration and Velocity at the Highest Point
At the highest point:
• Horizontal velocity component remains unchanged: same as initial vx.
• Vertical velocity component is 0 m/s because the projectile stops rising and is about to start falling.
• Horizontal acceleration is 0 m/s2 because no forces are acting in the horizontal direction.
• Vertical acceleration remains -9.8 m/s2 (negative because it is directed downwards).
Which of the three states of matter has the highest density?
A:solid
B:liquid
C:gas
Answers
Gas has the least density because the atoms and molecules spread to fit the container. Liquids is between solid and gas when it comes to density, while solid has the highest density due to the way the molecules are tightly and closely connected.
How does distance affect force when the amount of work remains the same?
A) As distance increases the amount of force needed increases.
B) As distance decreased the amount of force needed decreases.
C) As the distance increases the amount of force needed decreases.
D) As the distance increases the amount of force is not affected.
Answers
the way to represent a stable bond by using electron configuration is to have 7 electrons in the outer level? true or false ...?
Answers
The statement "The way to represent a stable bond by using electron configuration is to have 7 electrons in the outer level" is false since the magic number of electrons in a full level is 8.
Compare the lunar phase of the Waxing Crescent to the Waxing Gibbous.
A. The Waxing Crescent is increasing in illumination and the Waxing Gibbous is decreasing in illumination.
B. The Waxing Crescent is decreasing in illumination and the Waxing Gibbous is increasing in illumination.
C. The Waxing Crescent is more than half illuminated and the Waxing Gibbous is less than half illuminated.
D.The Waxing Crescent is less than half illuminated and the Waxing Gibbous is more than half illuminated.
Answers
By comparing the Waxing Crescent and Waxing Gibbous lunar phases, it can be seen that the Waxing Crescent is more illuminated while the Waxing Gibbous is less illuminated. Hence option A is correct.
What is Waxing Crescent ?The second stage of the cycle of phases is the Waxing Crescent. Once a month, this Moon phase lasts for 7.38 days before transitioning into the First Quarter phase. It rises at 9 AM and sets at 9 PM. The reason this phase is known as the Waxing Crescent is because the region of the Moon's surface that is lit resembles the shape of a crescent, and waxing refers to growth. The Earth, Moon, and Sun are practically perpendicular at this phase because it is one cycle away from the First Quarter phase. This indicates that the gravitational attraction of the tides from the Sun and Moon cancels out, resulting in a smaller tidal pull. At this time, the Earth's tides are practically at neap tide.
Hence option A is correct.
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The correct answer to compare the lunar phase of the Waxing Crescent to the Waxing Gibbous is D: The Waxing Crescent is less than half illuminated and the Waxing Gibbous is more than half illuminated.
Both phases are increasing in illumination as the visible portion of the moon grows. During the Waxing Crescent, there is a growing small portion, about 1/4, on the right side of the moon that is lit. As the moon moves towards the Waxing Gibbous phase, the illuminated portion increases to more than half, approximately 3/4, on the right side of the moon.
In both waxing phases, the angle formed by pointing one arm at the Moon and one arm at the Sun demonstrates an increase in the lit portion of the moon we see. In the Waxing Crescent phase, this angle is acute, and it becomes obtuse during the Waxing Gibbous phase, indicating the moon's journey towards a full moon, when the angle is 180°, and the entire near side of the moon, as viewed from Earth, is illuminated.
A student drops a rock from a bridge to the
water 12 m below.
With what speed does the rock strike
the water?
Answer in units of m/s ...?
Answers
a) In the first case g = 9.8 m/s^2 and S = 12 m.
So v = 14.21 m/s
Answer:
15.49m/s
Explanation:
Using one of the equations of motion;
[tex]v^{2}[/tex] = [tex]u^{2}[/tex] + 2gs
Where;
v = the final velocity of the rock.
=> This is the speed at which the rock strikes the water.
u = initial velocity of the rock.
=> The rock is just dropped from that height. That means the initial velocity is zero (0).
=> u = 0
g = the acceleration due to gravity.
=> Since the rock moves downwards, the acceleration due to gravity is positive and let it have a value of 10m/[tex]s^{2}[/tex]
=> g = 10m/[tex]s^{2}[/tex]
s = the distance covered by the rock
=> s = 12m
Substituting these values into the equation above gives
[tex]v^{2}[/tex] = [tex]0^{2}[/tex] + 2(10 x 12)
[tex]v^{2}[/tex] = 240
v = [tex]\sqrt{240}[/tex]
v = 15.49m/s
Therefore the velocity(speed) with the rock strikes the water is 15.49m/s
A spherical balloon is 40 ft in diameter and surrounded by air at 60°F and 29.92 in Hg abs.(a) If the balloon is filled with hydrogen at a temperature of 70°F and atmospheric pressure, what total load can it lift? (b) If it contains helium instead of hydrogen, other conditions remaining the same, what load can it lift? Answers (a) 2381 lb; (b) 2209 lb
Answers
in a transverse wave the peak of the wave is the crest and the lowest spot is the trough
Answers
If something is missing, then put it please otherwise no-one can help you!
Hope this helps!
In the 1950s, an experimental train, which had a mass of 2.50X10^4 kg, was powered across a level track by a jet engine that produced a thrust of 5.00X10^5 N for a distance of 509m.
a. Find the work done on the train
b. Find the change in Kinetic energy.
c. Find the final kinetic energy of the train if it started from rest.
d. Find the final speed of the train if there had been no fricion. ...?
Answers
Answer:
a. 2.545 *10^8 J
b. 2.545 *10^8 J
c. 2.545 *10^8 J
d. 141.688 m/s
Explanation:
Data:
mass, m = 2.5*10^4 kg
Force, F = 5.00*10^5 N
distance, d = 509 m
a. Work, W = F*d = 5.00*10^5 N * 509 m = 2.545 *10^8 J
b. Change in kinetic energy, ΔKE = W = 2.545 *10^8 J
c. Change in kinetic energy = final kinetic energy - initial kinetic energy
If the train started from rest, initial kinetic energy = 0
Then, final kinetic energy, KEf = ΔKE = 2.545 *10^8 J
d. From the definition of kinetic energy we can get the final speed (vf) as follows:
KEf = (1/2)*m*vf^2
vf = √(KEf*2/m)
vf = √(2.545 *10^8*2/2.5*10^4 )
vf = 141.688 m/s
Which of these best arranges the planets in the solar system in the increasing order of their diameters?
A. Mars → Venus → Earth → Mercury
B. Neptune → Saturn → Jupiter → Uranus
C. Mercury → Mars → Neptune → Uranus
D. Earth → Saturn → Mars → Mercury
Answers
Answer:
C. Mercury → Mars → Neptune → Uranus
Explanation:
There are 8 planets in the solar system. The smallest planet is mercury and the largest planet is Jupiter. There are 4 rocky planets and 4 gaseous planets.
The arrangement of planets in increasing of their diameter is:
Mercury
Mars
Venus
Earth
Neptune
Uranus
Saturn
Jupiter
Thus, from the given options, the correct sequence is:
C. Mercury → Mars → Neptune → Uranus
what two objects have stored energy? A. ball rolling on the ground B. a small rock sitting on top of a big rock C. a stretched rubber band D. a stone lying on the ground
Answers
Potential energy basically means something has the possibilty to change. A stretched rubber band could be released at any moment, which would then create kinetic energy. If the band wasn't stretched it wouldn't have any potential energy.
The correct statements are:
B. a small rock sitting on top of a big rock
As the rock is at a height with respect to ground it has potential Energy
and
C. a stretched rubber band
A stretched rubber band has elastic potential energy
The others are actually moving and hence would consist of Kinetic energy. Potential energy is stored in objects that do not move and are stationary.
Santa has lots of mixed up socks in his sack if he has 6 green socks 4 gold socks 8 black socks and 2 red socks what is the minimum he needs to pull out of his sack to get a matching pair
Answers
So, even if he's in a dark room and does it all without looking,
he will absolutely positively definitely have at least 1 matching
pair in has hand after he pulls out 5 socks.
The first 4 could have all been different colors. But the 5th sock
MUST match one of the other four.
pieces of rock that strike the surface of a moon or planet
Answers
What is pressure?
force per unit area
gravity per unit area
weight of the atmosphere
force per volume
Answers
Pressure is the force per unit area. It is calculated by dividing the force applied on an object by the area over which the force is distributed.
Explanation:Pressure is defined as force per unit area.
Pressure is calculated by dividing the force applied on an object by the area over which the force is distributed.
For example, if you push down on a table with a force of 50 Newtons over an area of 2 square meters, the pressure on the table is 25 Newtons per square meter.
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A girl and boy pull in opposite directions on a stuffed animal. The girl exerts a force of 3.5 N. The mass of the stuffed animal is 0.2 kg, and it is accelerating to the right 2.5 m/s2. What is the force that the boy is exerting?
Answers
2.5 x .2 = .5
3.5N + .5N = 4N
To find the force that the boy is exerting on the stuffed animal, subtract the net force from the girl's force. The boy is exerting a force of 3N.
Explanation:To find the force that the boy is exerting, we first need to determine the net force acting on the stuffed animal. The net force is equal to the mass of the stuffed animal multiplied by its acceleration. In this case, the mass is 0.2 kg and the acceleration is 2.5 m/s^2. So the net force is 0.2 kg * 2.5 m/s^2 = 0.5 N.
Since the girl exerts a force of 3.5 N, and the net force is 0.5 N, the boy's force must be equal to the difference between these two forces. Therefore, the boy is exerting a force of 3.5 N - 0.5 N = 3 N.
A tomato of mass 0.18 kg is dropped from a tall bridge. If the tomato has a speed of 11 m/s just before it hits the ground, what is the kinetic energy of the tomato?
Answers
K.E = 1/2 mv^2
K.E = 1/2 x 0.18 kg x 11 m/S^2
K.E = 0.99
Hope this helps
Julie carries an 8.0-kg suitcase as she walks 18 m along an inclined walkway to her hotel room at a constant speed of 1.5 m/s. The walkway is inclined 15 degrees above the horizontal. How much work does Julie do in carrying her suitcase?
Julie carries an 8.0-kg suitcase as she walks 18 m along an inclined walkway to her hotel room at a constant speed of 1.5 m/s. The walkway is inclined 15 degrees above the horizontal. How much work does Julie do in carrying her suitcase?
Answers
Answer:
[tex]W = 365.7 J[/tex]
Explanation:
As we know that work done is defined as the product of force and its displacement in the direction of the force
So here we know that
applied force by Julie is same as that of weight of her suitcase
So we will have
[tex]F = mg[/tex]
[tex]F = (8\times 9.81)[/tex]
[tex]F = 78.48 N[/tex]
now we know that the total height to which the suitcase is raised is given as
[tex]H = L sin\theta[/tex]
[tex]H = 18 sin15[/tex]
[tex]H = 4.66 m[/tex]
now the work done is given as
[tex]W = F.d[/tex]
[tex]W = 78.48(4.66)[/tex]
[tex]W = 365.7 J[/tex]