PHY183-Lecture14 - Physics for Scientists &...

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Unformatted text preview: Physics for Scientists & Engineers 1 Spring Semester 2011 Lecture 14 Power February 10, 2011 Physics for Scientists&Engineers 1 1 1 Clicker Quiz I pull a 2.0 kg mass and move it 2.0 m in the positive xdirection starting from x = 0. The force I exert is F ( x ) = ( 3.0 N/m 2 ) x 2 How much work have I done on the mass? • • • • • A) 0 J B) 4.0 J C) 6.0 J D) 8.0 J E) 16.0 J Physics for Scientists&Engineers 1 2 February 10, 2011 2 Clicker Quiz Solution W = ∫ F ( r ') • dr ' = ∫ F ( x ') dx ' F ( x ) = ( 3.0 N/m 2 ) x 2 2.0 m r0 0 r x W= ∫ 0 3.0 N/m 2 ) ( x ') dx ' ( 2 2 2.0 m 3.0 N/m 3 W = x 3 0 3.0 N/m 2 = (8.0 m3 ) = 8.0 J 3 D. February 10, 2011 Physics for Scientists&Engineers 1 3 3 Spring Force Equilibrium position x = x0 Stretched to x = x1 Stretched to x = x2 Compressed February 10, 2011 Physics for Scientists&Engineers 1 4 4 Spring Constant 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 to a length of 28.6 cm What is the value of the spring constant? February 10, 2011 Physics for Scientists&Engineers 1 5 5 Spring Constant We place the origin of our coordinate system so that x0 = -15.4 cm and x = -28.6 cm The spring force is Fs = −k ( x − x0 ) The force on the spring is provided by the weight of the 0.200 0.200 kg mass We can solve for k − (0.200 kg ) 9.81 m/s2 Fs mg k=− =− = = 14.9 N/m x − x0 x − x0 ( −0.286 m ) − ( −0.154 m ) ( ) February 10, 2011 Physics for Scientists&Engineers 1 6 6 Work Done by the Spring Force The work done by the spring in moving from x0 to x is WS = Fs ( x ') dx ' = x1 x2 x2 ( − kx ') dx ' = − k x2 x ' dx ' x1 x1 The work done by the spring is then 1 1 WS = − k x ' dx ' = − kx22 + kx12 2 2 x1 Taking x1 = x0 = 0 and x2 = x we get 1 WS = − kx2 2 February 10, 2011 x2 Physics for Scientists&Engineers 1 7 7 Compressing a Spring A massless spring located on a smooth horizontal surface is compressed by a force of 63.5 N, which results in a displacement of 4.35 cm from the initial equilibrium position. A steel ball of mass 0.075 kg is then placed in front of the spring and the spring is released. 2/10/2011 Physics for Scientists&Engineers 1 8 8 Compressing a Spring PROBLEM • What is the speed of the steel ball when it is shot off by the spring? THINK • If we compress a spring with an external force, we do work against the spring force • Releasing the spring by withdrawing the external force enables the spring to do work on the steel ball, which acquires kinetic energy in this process • Calculating the initial work done against the spring force enables us to figure out the kinetic energy that the steel ball will have and thus will lead us to the speed of the ball 2/10/2011 Physics for Scientists&Engineers 1 9 9 Compressing a Spring SKETCH • We draw a free-body diagram at the instant before the external force is removed • At this instant, the steel ball is at rest in equilibrium, because the external force and the spring force exactly balance each other • We set the x-coordinate of the ball at its left edge, which is where the ball touches the spring • This is the physically relevant location, because it measures the elongation of the spring from its equilibrium position 2/10/2011 Physics for Scientists&Engineers 1 10 10 Compressing a Spring RESEARCH • Once the external force is removed, the ball begins to move • The spring force is the only unbalanced force • The resulting acceleration is not constant • We will use energy concepts • Take x0 = 0 and initial compression of the spring to be xc Fs (xc ) = − kxc Fs (xc ) = − Fext kxc = Fext • We can now calculate k from the given information • Note that Fext < 0 and xc < 0 as we defined them Physics for Scientists&Engineers 1 2/10/2011 11 11 Compressing a Spring • We can now calculate the work necessary to compress the spring 12 kxc 2 • The work-kinetic energy theorem tells us W = −Ws = 12 W = ∆K K = K0 + W = 0 + kxc 2 • The kinetic energy of the ball is K= 12 mvx 2 2/10/2011 Physics for Scientists&Engineers 1 12 12 Compressing a Spring SIMPLIFY 2K vx = = m 2 12 kxc 2 m = 2 kxc = m Fext xc m CALCULATE vx = (−63.5 N )(−0.0435 m ) = 6.06877 m/s 0.075 kg ROUND vx = 6.1 m/s DOUBLE-CHECK • Units look good and speed seems reasonable 2/10/2011 Physics for Scientists&Engineers 1 13 13 Power Every car can get to the speed limit (in the US); but some cars get there quicker! Why? They have more power; and power costs money. Physics definition of power: power = rate at which work is done P= Average power dW dt W P= ∆t February 10, 2011 Physics for Scientists&Engineers 1 14 14 Power Unit Dimensional analysis: [ P ] = [W ] / [t ] = J/s (Work/time) Power received its own unit, the watt (W) • Possible confusion • italicized [W]: physical symbol for work • non-italicized W: symbol for the unit of power • Conversion: 1 W = 1 J/s 1 J = 1 kg m2/s3 • Useful everyday energy unit: 1 kWh = (1000 W)(3600 s) = 3.6 ⋅ 10 6 J = 3.6 MJ Non-SI power units (but widely used for cars) 1 hp = 746 W February 10, 2011 Physics for Scientists&Engineers 1 15 15 Clicker Quiz Which of the following is a correct unit for power? A. kg m/s2 B. N C. J D. m/s2 E. W 2/10/2011 Physics for Scientists&Engineers 1 16 16 Power for a Constant Force Work for a constant force: W = F • ∆r Differential: dW = F • dr (because F=constant) According to definition of power: dW P= dt F • dr = dt Final answer: Or: P = F •v P = Fv cos α Fv February 10, 2011 Physics for Scientists&Engineers 1 17 17 Accelerating Car Question: Car of mass 3,779 lb (= 1,714 kg) can reach the 60 mph (= 26.8 m/s) mark in 7.1 seconds. What is the average power needed to accomplish this? Answer: Car’s kinetic energy at 60 mph K = 1 mv 2 = 1 (1714 kg) ⋅ (26.8 m/s)2 = 616 kJ 2 2 Work-kinetic energy theorem: W = ∆K = K − K 0 = 616 kJ February 10, 2011 Physics for Scientists&Engineers 1 18 18 Accelerating Car Average power to get to the 60 mph mark in 7.1 seconds: W 6.16 ⋅ 10 5 J P= = = 86.8 kW=116 hp ∆t 7.1 s Discussion: Is this realistic? • No! 116 hp are not enough to accomplish this acceleration for a car of mass 1714 kg. Need at least 180 hp (this Aztek has 185 hp). • Why? Not perfectly efficient engine; not able to produce peak hp at all rpm; energy loss from friction and drag • Done Physics for Scientists&Engineers 1 February 10, 2011 19 19 Clicker Quiz A 1500 kg car accelerates from 0 to 25 m/s in 7.0 s. What is the average power delivered by the engine? Hint: 1 hp = 746 W A. 60 hp W ∆K P= = ∆t ∆t B. 70 hp 1 ∆K = mv 2 − 0 C. 80 hp 2 12 D. 90 hp mv mv 2 2 P= = E. 180 hp ∆t 2 ∆t (1500 kg )( 25 m/s)2 P= 2(7.0) s Physics for Scientists&Engineers 1 = 6.70 ⋅104 W = 90 hp 20 2/10/2011 20 Power to Keep a Car Moving A car travels at 60.0 mph on a level road. The car has a drag coefficient of cd =0.33 and a frontal area of 2.2 m2 . How much power does the car need to maintain its speed? Take the density of air to be 1.29 kg/m3. We can write the power as P = Fv = Fdrag v The force here is the force of air resistance Fdrag = Kv 2 K= 1 cd Aρair 2 We can combine these equations to get 1 1 P = cd Aρair v 2 v = cd Aρair v 3 2 2 So the power required to overcome air resistance is P= 1 3 0.33) ( 2.2 m 2 )(1.29 kg/m3 ) ( 26.8 m/s ) = 9.01 kW = 12.1 hp ( 2 Physics for Scientists&Engineers 1 21 February 10, 2011 21 Fuel Efficiency The average mass, power, and city-driving fuel efficiency of mid-sized cars sold in the US from 1975 to 2007 are shown here The mass of a car is important because of acceleration in stopand-go driving The The average power required to accelerate a car is proportional to the mass of the car 12 ∆W ∆K 2 mv mv2 P= = = = ∆t ∆t ∆t 2 ∆t 2/10/2011 Physics for Scientists&Engineers 1 22 22 Lifting Bricks PROBLEM • A load of bricks at a construction site has a mass of 85.0 kg. A crane raises this load from the ground to a height of 50.0 m in 60.0 s at a constant speed. What is the average power of the crane? THINK • Raising the bricks at a low constant speed means that the kinetic energy is negligible, so the work in this situation is done against gravity only • There is no acceleration, and friction is negligible • The average power then is just the work done against gravity divided by the time it takes to raise the load of bricks to the stated height SKETCH • In the free body diagram, the tension on the cable is equal to the weight of the bricks • The load is moved vertically a distance h 2/10/2011 Physics for Scientists&Engineers 1 23 23 Lifting Bricks RESEARCH • The work done by the crane is W = mgh • The average power required to lift the load in the given time is P= W ∆t SIMPLIFY SIMPLIFY P= CALCULATE mgh ∆t = 694.875 W P= (85.0 kg ) ( 9.81 m/s2 ) ( 50.0 m ) 60.0 s 2/10/2011 Physics for Scientists&Engineers 1 24 24 Lifting Bricks ROUND P = 695 W DOUBLE-CHECK • Convert out result for the required average power from watts to horsepower 1 hp P = ( 695 W ) = 0.932 hp 746 W • So a 1 hp motor would be sufficient to lift the 85.0 kg load vertically a distance of 50 m in 60 s 2/10/2011 Physics for Scientists&Engineers 1 25 25 ...
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