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**7.4 Work Done by a Varying Force**

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**Work Done by a Varying Force**

Assume that during a very small displacement, Dx, F is constant For that displacement, W ~ F Dx For all of the intervals,

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**Work Done by a Varying Force, cont**

Sum approaches a definite value: Therefore: (7.7) The work done is equal to the area under the curve!!

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**Example 7.7 Total Work Done from a Graph (Example 7.4 Text Book)**

The net work done by this force is the area under the curve W = Area under the Curve W = AR + AT W = (B)(h) + (B)(h)/2 = (4m)(5N) + (2m)(5N)/2 W = 20J + 5J = 25 J

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**Work Done By Multiple Forces**

If more than one force acts on a system and the system can be modeled as a particle, the total work done ON the system is the work done by the net force (7.8)

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**Work Done by Multiple Forces, cont.**

If the system cannot be modeled as a particle, then the total work is equal to the algebraic sum of the work done by the individual forces

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**Hooke’s Law The force exerted BY the spring is Fs = –kx (7.9)**

x is the position of the block with respect to the equilibrium position (x = 0) k is called the spring constant or force constant and measures the stiffness of the spring (Units: N/m) This is called Hooke’s Law

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Hooke’s Law, cont. When x is positive (spring is stretched), Fs is negative When x is 0 (at the equilibrium position), Fs is 0 When x is negative (spring is compressed), Fs is positive

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Hooke’s Law, final The force exerted by the spring (Fs ) is always directed opposite to the displacement from equilibrium Fs is called the restoring force If the block is released it will oscillate back and forth between –x and x

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**Work Done by a Spring Identify the block as the system**

The work as the block moves from: xi = – xmax to xf = 0 (7.10)

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**Work Done by a Spring, cont**

The work as the block moves from: xi = 0 to xf = xmax (7.10) (a)

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**Work Done by a Spring, final**

Therefore: Net Work done by the spring force as the block moves from –xmax to xmax is ZERO!!!! For any arbitrary displacement: xi to xf : (7.11)

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**Spring with an Applied Force**

Suppose an external agent, Fapp, stretches the spring The applied force is equal and opposite to the spring force: Fapp = –Fs = –(–kx) Fapp = kx

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**Spring with an Applied Force, final**

Work done by Fapp when xi = 0 to xf = xmax is: WFapp = ½kx2max For any arbitrary displacement: xi to xf : (7.12)

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Active Figure 7.10

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**7.5 Kinetic Energy And the Work-Kinetic Energy Theorem**

Kinetic Energy is the energy of a particle due to its motion K = ½ mv (7.15) K is the kinetic energy m is the mass of the particle v is the speed of the particle Units of K: Joules (J) 1 J = N•m = (kg•m/s2)m= kg•m2/s2 = kg(m/s)2 A change in kinetic energy is one possible result of doing work to transfer energy into a system

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**Kinetic Energy, cont Calculating the work: Knowing that:**

F = ma = mdv/dt =m(dv/dt)(dx/dx) Fdx = m(dv/dx)(dx/dt)dx = mvdv (7.14)

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**Work-Kinetic Energy Theorem**

The Work-Kinetic Energy Principle states SW = Kf – Ki = DK (7.16) In the case in which work is done on a system and the only change in the system is in its speed, the work done by the net force equals the change in kinetic energy of the system. We can also define the kinetic energy K = ½ mv (7.15)

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**Work-Kinetic Energy Theorem, cont**

Summary: Net work done by a constant force in accelerating an object of mass m from v1 to v2 is: Wnet = ½mv22 – ½mv12 DK “Net work on an object = Change in Kinetic Energy” It’s been shown for a one-dimension constant force. However, this is valid in general!!!

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**Work-Kinetic Energy REMARKS!!**

Wnet ≡ work done by the net (total) force. Wnet is a scalar. Wnet can be positive or negative since K can be both + or – K ½mv2 is always positive. Mass and v2 are both positive. (Question 10 Homework) Units are Joules for both work & kinetic energy. The work-kinetic theorem: relates work to a change in speed of an object, not to a change in its velocity.

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**Example 7.8 Question #14 (a). Ki ½m v2 ≥ 0**

K depends on: v2 ≥ 0 & m > 0 If v 2v Kf = ½m (2v)2 = 4(½mv2 ) = 4Ki Then: Doubling the speed makes an object’s kinetic energy four times larger (b). If SW = 0 v must be the same at the final point as it was at the initial point

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**Example 7.9 Work-Kinetic Energy Theorem (Example 7.7 Text Book)**

m = 6.0kg first at rest is pulled to the right with a force F = 12N (frictionless). Find v after m moves 3.0m Solution: The normal and gravitational forces do no work since they are perpendicular to the direction of the displacement W = F Dx = (12)(3)J = 36J W = DK = ½ mvf2 – 0 36J = ½(6.0kg)vf2 = (3kg)vf2 Vf =(36J/3kg)½ = 3.5m/s

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**Example 7.10 Work to Stop a Car**

Wnet = Fdcos180°= –Fd = –Fd Wnet = K = ½mv22 – ½mv12 = –Fd -Fd = 0 – ½m v12 d v12 If the car’s initial speed doubled, the stopping distance is 4 times greater. Then: d = 80 m

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**Example 7.11 Moving Hammer can do Work on Nail**

A moving hammer strikes a nail and comes to rest. The hammer exerts a force F on the nail, the nail exerts a force –F on the hammer (Newton's 3rd Law) m Work done on the nail is positive: Wn = Kn = Fd = ½mnvn2 – 0 > 0 Work done on the hammer is negative: Wh = Kh = –Fd = 0 – ½mhvh2 < 0

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**Example 7.12 Work on a Car to Increase Kinetic Energy**

Find Wnet to accelerate the 1000 kg car. Wnet = K = K2 – K1 = ½m v22 – ½m v12 Wnet = ½(103kg)(30m/s)2 – ½(103kg)(20m/s)2 Wnet = 450,000J – 200,000J = 2.50x105J

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**Example 7.13 Work and Kinetic Energy on a Baseball**

A 145-g baseball is thrown so that acquires a speed of 25m/s. ( Remember: v1 = 0) Find: (a). Its K. (b). Wnet on the ball by the pitcher. (a). K ½mv2 = ½(0.145kg)(25m/s)2 K 45.0 J (b). Wnet = K = K2 – K1 = 45.0J – 0J Wnet = 45.0 J

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7.6 Non-isolated System A nonisolated system is one that interacts with or is influenced by its environment An isolated system would not interact with its environment The Work-Kinetic Energy Theorem can be applied to nonisolated systems

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Internal Energy The energy associated with an object’s temperature is called its internal energy, Eint In this example, the surface is the system The friction does work and increases the internal energy of the surface

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Active Figure 7.16

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Potential Energy Potential energy is energy related to the configuration of a system in which the components of the system interact by forces Examples include: elastic potential energy – stored in a spring gravitational potential energy electrical potential energy

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**Ways to Transfer Energy Into or Out of A System**

Work – transfers by applying a force and causing a displacement of the point of application of the force Mechanical Waves – allow a disturbance to propagate through a medium Heat – is driven by a temperature difference between two regions in space

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**More Ways to Transfer Energy Into or Out of A System**

Matter Transfer – matter physically crosses the boundary of the system, carrying energy with it Electrical Transmission – transfer is by electric current Electromagnetic Radiation – energy is transferred by electromagnetic waves

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**Examples of Ways to Transfer Energy**

a) Work b) Mechanical Waves c) Heat d) Matter transfer e) Electrical Transmission f) Electromagnetic radiation

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**Conservation of Energy**

Energy is conserved This means that energy cannot be created or destroyed If the total amount of energy in a system changes, it can only be due to the fact that energy has crossed the boundary of the system by some method of energy transfer Mathematically: SEsystem = ST (7.17) Esystem is the total energy of the system T is the energy transferred across the system boundary Established symbols: Twork = W and Theat = Q The Work-Kinetic Energy theorem is a special case of Conservation of Energy

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**Material for the Final Examples to Read!!! Example 7.7 (page 195)**

Homework to be solved in Class!!! Problems: 11, 26

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Energy and Energy Transfer

Energy and Energy Transfer

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