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### PHY183-Lecture14pre

Course: PHY 183, Spring 2012
School: Michigan State University
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Word Count: 2074

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is What Energy? Richard Feynman talks about the concept of energy: It is important to realize that in physics today, we have no knowledge of what energy is. We do not have a picture that energy comes in little blobs of a definite amount. It is not that way. However, there are formulas for calculating some numerical quantity, and when we add it all together it gives 28always the same number. It is an abstract...

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is What Energy? Richard Feynman talks about the concept of energy: It is important to realize that in physics today, we have no knowledge of what energy is. We do not have a picture that energy comes in little blobs of a definite amount. It is not that way. However, there are formulas for calculating some numerical quantity, and when we add it all together it gives 28always the same number. It is an abstract thing in that it does not tell us the mechanism or the reasons for the various formulas. Physics for Scientists & Engineers 1 Fall Semester 2012 Lecture 12 Energy, Kinetic Energy, Work February 7, 2012 Physics for Scientists&Engineers 1 1 K car = 1 mv 2 = 1 (1310 kg)(24.6 m/s)2 = 4.0 10 5 J 2 2 Mass of the Earth is 61024 kg, and it orbits the Sun with a speed of 30,000 m/s 2 Unit of kinetic energy: [ K ] = [ m ] [ v]2 = kg m 2 / s 2 K sun = 1 mv 2 = 1 (6.0 10 24 kg)(3.0 10 4 m/s)2 = 2.7 10 33 J 2 2 Baseball (mass 5 ounces = 0.142 kg) thrown at 80 mph (= 35.8 m/s) has kinetic energy This energy unit has received its own name, Joule (J), named after British physicist James Joule (1818-1889) Energy unit 1 J = 1 kg m 2 / s2 Useful conversion: 1 J = 1 N m K baseball = 1 mv 2 = 1 (0.142 kg)(35.8 m/s)2 = 91 J 2 2 Electron (me = 9.110-31 kg) moving with a speed of 1.3106 m/s (= 0.4% of the speed of light): K e = 1 mv 2 = 1 (9.1 10 31 kg)(1.3 10 6 m/s)2 = 7.7 10 19 J 2 2 3 February 7, 2012 Other Energy Units 4 Question: A crystal vase (mass 2.40 kg) is dropped from a height of 1.30 m and falls to the floor. What is its kinetic energy just before impact? Answer: Once we know the velocity of the vase just before impact, we can put it into our equation for the kinetic energy. To obtain this velocity, we remind ourselves of the kinematics of free-falling objects. 2 Use: v2 = v0 2 g( y y0 ) = 2 g( y0 y ) ( v0 = 0 ) Now use definition of kinetic energy and obtain in this case: 1 eV = 1.60210-19 J (electron in previous example has ~5 eV kinetic energy) The energy you eat is measured in food calories 1 Cal = 4186 J (1 Cal > 40 times the kinetic energy of the baseball in previous example) For very large energy measures one need a large unit. The energy released by one million tons of TNT is large 1 Mt = 4.01015 J Physics for Scientists&Engineers 1 Physics for Scientists&Engineers 1 Example: Falling Vase Atomic and nuclear physics introduced the electron-Volt February 7, 2012 2 Car of mass 1,310 kg driving 55 mph (=24.6 m/s) Energy contained in the motion of an object Definition K = 1 mv 2 Physics for Scientists&Engineers 1 Physics for Scientists&Engineers 1 Kinetic Energy Examples Kinetic Energy February 7, 2012 February 7, 2012 K = 1 mv2 = 1 m ( 2 g( y0 y)) = mg( y0 y) 2 2 Numbers: K = (2.40 kg) (9.81 m/s2 ) (1.30 m 0) = 30.6 J 5 February 7, 2012 Physics for Scientists&Engineers 1 6 1 Work Example: Falling Vase - Observations In the case of the falling vase, the kinetic energy is a function of the height (linearly dependent on height) from which the vase was released K = mg( y0 y ) Gravitational force, Fg = mg, accelerated vase during free fall, and thus increased its kinetic energy Kinetic energy is proportional to the magnitude of the gravitational force Here the kinetic energy gained by the vase is simply the product of the magnitude of the gravitational force and the displacement February 7, 2012 Physics for Scientists&Engineers 1 7 Work Done by a Constant Force In the previous example, the vase started out with zero kinetic energy and after falling it had gained a kinetic energy of 30.6 J The gravitational force accelerates the vase and gives it kinetic energy We account for this change in the kinetic energy of an object caused by a force with the concept of work, W Formal Definition of Work: Work is the energy transferred to or from an object due to the action of a force. Positive work is a transfer of energy to the object, and negative work is a transfer of energy from the object. February 7, 2012 Physics for Scientists&Engineers 1 8 Work Done by a Constant Force Suppose we let the vase in the previous example slide from rest along an inclined plane that has an angle with respect to the horizontal In our snowboarding example we showed that a = g sin In this example we have = 90 so a = g cos We can determine the kinetic energy of the vase as a function of the displacement using kinematics 2 v2 = v0 + 2ar taking v0 = 0 v2 = 2 g cos r 1 K = mv2 = mg r cos 2 February 7, 2012 Physics for Scientists&Engineers 1 9 Work Done by a Constant Force Physics for Scientists&Engineers 1 10 Work Done by a Constant Force W = Fr Because Fg = mg The kinetic energy transferred to the vase was the result of positive work done by the force of gravity so W = F r cos , where is the angle between F and r K = mg r cos = Wg Limiting cases February 7, 2012 W = Fr cos This equation for the work done by a constant force acting over some spatial displacement holds for all constant force vectors and arbitrary displacement vectors W =0 = 0 : Fg and r are parallel and pointing downward Wg = mgr = 90 : Fg points downward but the vase cannot move downward Wg = 0 February 7, 2012 Physics for Scientists&Engineers 1 11 February 7, 2012 Physics for Scientists&Engineers 1 12 2 Work Done by Constant Force - Example Work as a Scalar Product Using a scalar product, we can write the work done by a constant force as W = F r W = F r cos = Fx x + Fy y + Fz z ) ( 13 )( )( )( ) February 7, 2012 Physics for Scientists&Engineers 1 14 Equivalent formulations (solving for K or K0): K = K0 + W K0 = K W K K K 0 = W This is the work-kinetic energy theorem 1212 mvx mvx 0 = max ( x x0 ) = Fx x = W 2 2 The work-kinetic theorem is equivalent to Newtons Second Law 15 Work Done by Gravity Kinetic energy cannot be less than zero The work that can be done by an object (negative work) cannot have a bigger magnitude than the initial kinetic energy The work-kinetic energy theorem is valid for constant forces and variable forces February 7, 2012 Physics for Scientists&Engineers 1 16 Work Done Lifting an Object Application of an external force F in vertical direction Work-kinetic energy theorem now includes work done by gravity and work done by external force Work done by gravitational force during upward motion of a projectile Wg = -mgh K K0 = Wg + WF (reduces energy kinetic on the way up) When we lift an object, it is initially at rest (K0 = 0) and also after we lifted it to its final height (K = 0) => W = W Work done by gravitational force on the projectile on the way down Wg = mgh F g Lifting: WF = -Wg = mgh Lowering: WF = -Wg = -mgh (increases kinetic energy on the way down) Physics for Scientists&Engineers 1 )( Work-Kinetic Energy Theorem Relationship between kinetic energy of an object and the work done on the object by a constant force February 7, 2012 )( W = 4.79 N 4.25 m + 3.79 N 3.69 m + 2.09 N 2.45 m W = 1.25 J Work-Kinetic Energy Theorem Physics for Scientists&Engineers 1 W = F r = Fx rx + Fy ry + Fz rz W = F r = Fx x = Fx ( x x0 ) February 7, 2012 ) The work is given by In one dimension Physics for Scientists&Engineers 1 ) object of mass 18.0 kg, causing a displacement of that object by r = 4.25,3.69, 2.45 m. What is the total work done by ( Wnet = Fnet r = Fi r = Fi r = Wi i i i February 7, 2012 ( A constant force, F = 4.79, 3.79,2.09 N acts on an this force? The net work done by a net force is equal to the sum of the work done by the individual forces ( PROBLEM 17 February 7, 2012 Physics for Scientists&Engineers 1 18 3 Example: Weightlifting Lifting with Pulleys Question: Ronny Weller won silver in Sidney in 2000. He lifted 257.5 kg in the jerk competition. Assuming he lifted weight to height of 1.83 m and held it there, what was the mechanical work done by him in this process? Answer: The work is W = mgh = (257.5 kg) (9.81 m/s ) (1.83 m) 2 19 Work Done by Variable Force However, displacements are different: r1 = 2r2 So we get for work done by T1: x x0 The work done can be represented by the area under the curve of Fx(x) vs. x 21 Work Done by a Variable Force - Example xi Corresponds to the area under the curve in our sketch SIMPLIFY 4.0 Physics for Scientists&Engineers 1 22 The units are good. Lets calculate the area under the curve triangle rectangle 1 W = ( 3.0 N ) ( 4.0 m ) + ( 4.0 m ) ( 2.0 N ) 2 W = 12 J + 4.0 J = 16 J xf 0.0 February 7, 2012 DOUBLE-CHECK W = F ( x ) dx ( 3.0 + 0.50 x ) dx = [ 3.0 x ] 20 Work Done by a Variable Force - Example RESEARCH The work done by a variable force is given by 4.0 Identical work with or without pulleys! Physics for Scientists&Engineers 1 Work Done by Variable Force - Example W = Fx ( x ') dx ' Physics for Scientists&Engineers 1 February 7, 2012 A particle of mass m is subjected to a force acting in the x-direction, Fx = (3.0+0.50x) N. PROBLEM Find the work done as the particle moves from x = 0.0 to x = 4.0 m. THINK This forces varies with x, so we need to integrate over the displacement. SKETCH Suppose the force acting on an object is not constant The work done by such a force is W= W2 = T2 r2 = T2 r2 = mgr2 = mgr2 = W2 Physics for Scientists&Engineers 1 February 7, 2012 T1 = 1 T2 = 1 mg 2 2 Work done by T2: W1 = T1 r1 = ( 1 T2 )(2r2 ) 2 = 4.62 kJ Note: The work in lowering is then -4.62 kJ. February 7, 2012 Force needed to pull mass m up with pulleys is half of weight 4.0 + 0.25 x2 0.0 0.0 W = ( 3) ( 4.0 ) + ( 0.25 ) (16 ) = 16 J ROUND W = 16 J February 7, 2012 Physics for Scientists&Engineers 1 23 February 7, 2012 Physics for Scientists&Engineers 1 24 4 The Spring Force Work-Kinetic Energy Theorem for Variable Force Assume a variable force in the x-direction A very common variable force is the spring force We can treat a spring in terms of motion and forces in one dimension Fx ( x ) = ma We can use the chain rule to get dv dv dx = dt dx dt Integrate over the displacement to get the work Choose the x-direction a= x x x x0 x0 x0 W = Fx ( x ') dx ' = ma dx ' = m We will use the spring force extensively throughout the semester So lets get a strong foundation now dv dx ' dx ' dx ' dt The spring force occurs in many physical situations We start with a spring that is neither stretched nor compressed Change the variable of integration and do the integral x W = m x0 v v v '2 dv dx ' dx ' = mv ' dv ' = m = K K0 = K dx ' dt 2 v0 v0 February 7, 2012 Physics for Scientists&Engineers 1 Equilibrium position of the end of the spring, x0 25 February 7, 2012 Spring Force Physics for Scientists&Engineers 1 26 Spring Spring Force We can summarize by stating Equilibrium position x = x0 The magnitude of the spring force is proportional to the absolute value of the displacement from the equilibrium position The spring force is always in the opposite direction from the displacement Stretched to x = x1 Stretched to x = x2 Fs = k ( x x0 ) k is the spring constant Always positive by definition Has units of N/m Compressed The negative sign means that the spring force is always in the opposite direction from the displacement February 7, 2012 Physics for Scientists&Engineers 1 27 Spring Force Fs = kx Hookes Law Named after British physicist Robert Hooke (1635 1703), contemporary of Newton The spring force always points toward the equilibrium position Restoring force Linear restoring forces are found in many systems in nature Crystal lattice Shape deformations in nuclei . Physics for Scientists&Engineers 1 Physics for Scientists&Engineers 1 28 Spring Constant If we take x0 = 0, we have February 7, 2012 February 7, 2012 29 A spring has a length of 15.4 cm and is hanging vertically from a support point above it A weight with a mass of 0.200 kg is attached to the spring, causing it to extend to a length of 28.6 cm What is the value of the spring constant? February 7, 2012 Physics for Scientists&Engineers 1 30 5 Spring Constant Work Done by the Spring Force We place the origin of our coordinate system so that x0 = -15.4 cm and x = -28.6 cm The spring force is The work done by the spring in moving from x0 to x is x Physics for Scientists&Engineers 1 x0 x 1 12 WS = k x ' dx ' = kx2 kx0 2 2 x0 ) Taking x0 = 0 we get ( 0.200 kg ) 9.81 m/s2 Fs mg = = = 14.9 N/m x x0 x x0 0.286 m ) ( 0.154 m ) ( February 7, 2012 x0 The work done by the spring is then The force on the spring is provided by the weight of the 0.200 kg mass We can solve for k ( x x0 Fs = k ( x x0 ) k= x WS = Fs ( x ') dx ' = ( kx ') dx ' = k x ' dx ' 1 WS = kx2 2 31 February 7, 2012 Physics for Scientists&Engineers 1 32 6
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Michigan State University - PHY - 183
Michigan State University - PHY - 183
Michigan State University - PHY - 183
Michigan State University - PHY - 183
Michigan State University - PHY - 183
Michigan State University - PHY - 183
Michigan State University - PHY - 183
Michigan State University - PHY - 183
Michigan State University - PHY - 183
Michigan State University - PHY - 183
Michigan State University - PHY - 183
Michigan State University - PHY - 183
Michigan State University - PHY - 183
Michigan State University - PHY - 183
Michigan State University - PHY - 183
Michigan State University - PHY - 183
Michigan State University - PHY - 183
Michigan State University - PHY - 183
Michigan State University - PHY - 183
Michigan State University - PHY - 183
Michigan State University - PHY - 183
Michigan State University - PHY - 183
Michigan State University - PHY - 183
Michigan State University - PHY - 183
Michigan State University - PHY - 183
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Michigan State University - PHY - 183
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