The pulling force is F-111 N parallel to the incline, which makes an angle of 19.5" with the horizontal. The coefficient of kinetic friction is 0.400, and the crate is pulled d 5.11 m A crate of mass m -9.7 (a) How much work is done by the gravitational force on the crate? (b) Determine the increase in internal energy of the crate-incline system owing to friction (c) How much work is done by the 111-N force on the crate (d) What is the change in kinetic energy of the crate? (e) What is the speed of the crate after being pulled 5.11 m
Kinematics
A machine is a device that accepts energy in some available form and utilizes it to do a type of work. Energy, work, or power has to be transferred from one mechanical part to another to run a machine. While the transfer of energy between two machine parts, those two parts experience a relative motion with each other. Studying such relative motions is termed kinematics.
Kinetic Energy and Work-Energy Theorem
In physics, work is the product of the net force in direction of the displacement and the magnitude of this displacement or it can also be defined as the energy transfer of an object when it is moved for a distance due to the forces acting on it in the direction of displacement and perpendicular to the displacement which is called the normal force. Energy is the capacity of any object doing work. The SI unit of work is joule and energy is Joule. This principle follows the second law of Newton's law of motion where the net force causes the acceleration of an object. The force of gravity which is downward force and the normal force acting on an object which is perpendicular to the object are equal in magnitude but opposite to the direction, so while determining the net force, these two components cancel out. The net force is the horizontal component of the force and in our explanation, we consider everything as frictionless surface since friction should also be calculated while called the work-energy component of the object. The two most basics of energy classification are potential energy and kinetic energy. There are various kinds of kinetic energy like chemical, mechanical, thermal, nuclear, electrical, radiant energy, and so on. The work is done when there is a change in energy and it mainly depends on the application of force and movement of the object. Let us say how much work is needed to lift a 5kg ball 5m high. Work is mathematically represented as Force ×Displacement. So it will be 5kg times the gravitational constant on earth and the distance moved by the object. Wnet=Fnet times Displacement.
![Part 6 of 7 - Analyze
(d) We use the energy version of the nonisolated system model to find the change in kinetic energy.
AK
Wother AEint
=WF + Wg - AE int
567
J+
The response you submitted has the wrong sign. ]
183
222
J
Note that the normal force does zero work because it is at
x
Your response differs from the correct answer by more than 10%. Double check your calculations.° to the motion.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9019e6de-aeac-44f2-b3de-2e9ccb012f8d%2F978f947e-26e6-4ce1-bb63-e7d1b4d4f3ed%2Fvmspvj_processed.jpeg&w=3840&q=75)
![A crate of mass m -9.7 kg is pulled up a rough incline with an initial speed of v, 1.43 m/s. The pulling force is F-111 N parallel to the incline, which makes an angle of 0 19.5" with the horizontal. The coefficient of kinetic friction is 0.400, and the crate is pulled d=5.11 m.
(a) How much work is done by the gravitational force on the crate?
(b) Determine the increase in internal energy of the crate-incline system owing to friction.
(c) How much work is done by the 111-N force on the crate?
(d) What is the change in kinetic energy of the crate?
(e) What is the speed of the crate after being pulled 5.11 m?
Part 1 of 7 Conceptualize
The gravitational force does negative work of some tens of joules on the crate. We expect some hundreds of joules of work done by the force on the crate. This force should be larger than the increase in the internal energy of the system and larger than the change in kinetic energy of the crate. We estimate the final speed to be a few meters per second.
Part 2 of 7 Categorize
We could use Newton's second law to find the crate's acceleration, but using ideas of work and energy is a more direct way to calculate the final speed. Finding the increase in internal energy is a step towards finding the temperature increase of the rubbing surfaces.
Part 3 of 7-Analyze
(a) The force of gravitation is (9.7 kg)(9.80 m/s²)-95.1 N straight down, at an angle of (90✔
-(95.1 N) 5.11
5.11 m) cos 109.5
110
90 19.5) 109.5
110 to the motion. The work done by gravity on the crate is given by
Part 4 of 7-Analyze
(b) We set the x and y axes parallel and perpendicular to the incline. From Newton's second law, we know that
and we have
(95.1 N) cos 19.5✔
19.5 son 9.7✔
89.6 N.
Substituting for the coefficient of friction and the normal force,
0.4
89.6
Therefore, for the change in internal energy we have
35.9✔
35.9 (5.11✔
Part 5 of 7-Analyze
(c) For the work done by the force on the crate, we have
W-F
5.11 m) cos✔
35.9 N.
5.11 m)-183.35
183 J.
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