A Driver In A 1000 Kg Car Traveling – 6.2 Kinetic Energy and the Work-Energy Theorem 6.3 Work and Energy with Various Forces 6.4 Force October 7-13, 2013
The energy approach to describing motion is particularly useful when Newton’s Laws are difficult or impossible to apply. Energy is a scalar quantity. There is no direction associated with it. October 7-13, 2013
A Driver In A 1000 Kg Car Traveling
4 Kinetic energy Kinetic energy is the energy associated with the object’s state of motion v For an object moving with a velocity in SI units: joule (J) 1 joule = 1 J = 1 kg m2/s2 October 7-13, 2013
A 1000 Kg Car Is Travelling At 108 Km/h When The Brakes Are Applied. If The Car Stops In 10 M, Find The Stopping Force
Work provides the relationship between force and energy Work done on an object is transferred to/from it If W > 0, energy added: “transferred to the object” If W < 0, energy removed: "transferred from the object" October 7 – 13, 2013
8 Definition of work W, the work done on an object by a constant force, is defined as the product of the force component in the direction of displacement and the magnitude of the displacement F Δ x the magnitude of the force. The magnitude of the object’s displacement q is the angle between October 7 and 13, 2013
Work required to change the speed or acceleration of an object is a scalar number SI unit Newton • meter = Joule N • m = J J = kg • m2 / s2 = (kg • m/s2) • m October 7-13, 2013
10 Action: + or -? The function can be positive, negative, or zero. The sign of the work depends on the direction of the force relative to the displacement. = 180° October 7-13, 2013
Answered: 7. A Car (mass 1000 Kg) Is Traveling At…
A man carries a bucket of water horizontally at a constant speed. Force does not act on bucket Transfer horizontal force vertical cos 90° = 0 October 7-13, 2013
When the frame is raised the work is positive, if the box is lowered the work would be negative The force would still go up but the displacement would be lower October 7-13, 2013
The work W done by an agent exerting a constant force on the system is the product of the magnitude of the force F, the magnitude of the displacement Δρ of the point of application of the force, and cosθ, where θ is the angle. between force and displacement vectors: I II IV III October 7-13, 2013
14 Work and Power Return An Eskimo pulls a sled as shown. The sled has a total mass of 50.0 kg, and it exerts a force of 1.20 × 102 N on the sled pulling the rope. How much work is done on the sled if θ = 30° and the sled pulls 5.0 m? October 7-13, 2013
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If more than one force acts on an object, the total work is equal to the algebraic sum of the work done by the individual forces. Remember that work is a scalar, and an algebraic sum is also October 7-13, 2013.
18 The Work-Energy Theorem When work is done on an object by a net force and the only change in the object is its velocity, the work done is equal to the change in the object’s kinetic energy, and if the work is positive, the velocity increases. . if the case is negative, October 7-13, 2013
Determine the initial and final positions of the body, and draw a free-body diagram showing and labeling all the forces acting on the body, choose a coordinate system, list the unknowns and knowns, and the unknowns determine your target variables Calculate the work done. with every effort. Check the signs. Add the amount of work done by each force to find the net (total) work Wnet October 7-13, 2013 Check the meaning of your answer.
A driver of a 1.00-103-kg car slams on his brakes at 35.0 m/s on an interstate to avoid hitting a second car that is at rest due to traffic ahead. After the brakes are applied, a constant frictional force of 8.00-103 N acts on the car. Ignore air resistance. (a) At what minimum distance must the brakes be applied to avoid a collision with another vehicle? (b) If the vehicles are initially only m apart, at what speed will the collisions occur? October 7-13, 2013
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(b) We know that find the velocity at impact. Write the work-energy theorem: At a given distance of 30.0 m, a car is very close to another car. October 7-13, 2013
Force versus stance position The total work done by the force is represented by the area under the curve between the starting position and the final position.
Work energy theorem Wtol = K for varying forces as well as for constant forces October 7-13, 2013
This includes the spring constant, k Hooke’s law directs the force F in the opposite direction to x, always returning to the equilibrium point. k depends on how the spring is made, material, wire thickness, etc. Unit: N/m. October 7-13, 2013
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Begin by springing at the length of your natural balance. Just a mass in the spring, let it be placed at a distance (stationary) From this you can get a permanent spring. December 8, October 7-13, 2018
We apply equal and opposite forces to the end of the spring and gradually increase the forces The work we must do to stretch the spring from x1 to x2 The work done on the spring is not the same as the work done on the spring 7 -13. October, 2013
The speed of energy transfer is important in practical device design and use, and the time of energy transfer is called speed. Medium energy is delivered by energy transfer, October 7-13, 2013
29 Momentary power is the time of energy transfer. A force is used for any means of transferring energy Another Statement October 7-13, 2013 General Definition of Momentary Force
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1 watt = 1 joule/second = 1 kg. m2/s3 The unit of power in the US customary system is the horsepower. 1 horsepower = 550 feet. f/s = 746 W Units of energy can also be used to express units of work or energy 1 kW = (1000 W)(3600 s) = 3.6 x106 J October 7-13, 2013
A 1000 kg lift increases the maximum load to 800 kg. A constant frictional force of 4000 N stops its upward motion. What is the minimum power required by the motor to raise the elevator at a constant speed of 3 m/s? October 7-13, 2013
To operate this website, we collect user data and share it with processors. To use this website, you must agree to our Privacy Policy, including our cookie policy. SciencePhysicsPROBLEM How to prevent a 1.00×10*3 kg driver from hitting the brakes on an interstate at 35.0 m/s (80.0 mph). the second car coming to rest due to the congestion in front of it (Figure 2.7). After the brake is applied, a kinetic friction force of magnitude 8.00×10*3 N acts on the car. Ignore air resistance. (a) At what minimum distance must the brakes be applied to avoid a collision with another vehicle? (b) If the vehicles are initially 30.0 m apart, at what speed will the collision occur?
PROBLEM A driver of a 1.00 x 10 x 3 kg car slams on his brakes on an interstate at 35.0 m/s (80.0 mph) before coming to rest to avoid hitting another car ahead. congestion (Figure 2.7). After the brake is applied, a kinetic friction force of magnitude 8.00×10*3 N acts on the car. Ignore air resistance. (a) At what minimum distance must the brakes be applied to avoid a collision with another vehicle? (b) If the vehicles are initially 30.0 m apart, at what speed will the collision occur?
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Transcription Image Text: Physics. Second chapter lecture No. 2 Prof. Dr. Jamal A. Example, Page 1 I EXAMPLE 2.2 Collision Analysis OBJECTIVE Apply the work-energy theorem to a given force. PROBLEM A driver of a 1.00 x 10 x 3 kg car slams on his brakes on an interstate at 35.0 m/s (80.0 mph) before coming to rest to avoid hitting another car ahead. congestion (Figure 2.7). After the brake is applied, a kinetic friction force of magnitude 8.00×10*3 N acts on the car. Ignore air resistance. (a) At what minimum distance must the brakes be applied to avoid a collision with another vehicle? (b) If the cars were initially 30.0 m apart, what would be their speed?
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