Créer une présentation
Télécharger la présentation

Télécharger la présentation
## Examples and Hints in Chapter 6

- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -

**6.41**• A small glider is placed against a compressed spring at the bottom of an air track that slopes upward at an angle of 400 above the horizontal. The glider has a mass of 0.09 kg. The spring has a spring constant, k=640 N/m and negligible mass. When the spring is released, the glider travels a maximum distance of 1.8 m along the air track before sliding back down. Before reaching this maximum distance, the glider loses contact with the spring. • A) What distance was the spring originally compressed? • B) When the glider has traveled along the air track 0.8 m from its initial position again the compressed spring, is it still in contact with the spring? What is the kinetic energy of the glider at this point?**Things to note**• Air track– No friction • Work done by spring is counteracted by the work done by gravity. • Work done by gravity can be calculated by F*distance*cosine(angle between them). mg L W= -mg*L*cos(500) W= -0.09*9.8*1.8*.647 W= -1.02 J 900-400 400**DK=Wtotal**• In this case, DK=0 so Wtotal=0 • Wtotal=Wspring+ Wgravity • Wtotal= ½ kX2 -1.02 J=0 • X2=1.02*2/640=0.031 m2 • X=0.056 m**Part B**• Wtotal= ½ mv2-0=½ mv2=K • Wgravity=-0.09*9.8*0.80*cos(500)=-0.45 J • Wspring= ½ k X2 =0.5*640*.0562=1.003 J • Wtotal= 1.003-0.45=0.5499 J=K • No longer in contact with spring after 5.6 cm**6.69**A small block with a mass of 0.12 kg is attached to a cord passing through a hole in a frictionless, horizontal surface. The block is originally revolving at a distance of 0.4 m from the hole with a speed of 0.70 m/s . The cord is then pulled from below, shortening the radius of the circle in which the block revolves to 0.10 m. At this new distance, the speed of the block is observed to be 2.8 m/s. • What is the tension in the cord in the original situation when the block has a speed of 0.7 m/s? • What is the tension in the cord in the final situation when the speed of the block is 2.8 m/s? • How much work was done by the person who pulled on the cord?**Freebody Diagrams**Part a) If m=0.12 kg and v=0.7 m/s and R=0.4 m then T must be 0.15 N Part b) If v=2.8 m/s and R=0.1 m then T must be 9.41 N T mv2/R**Part C)**• DK=W so • K2= ½ mv22 = 0.5 * 0.12kg*2.8^2=0.47 J • K1= ½ mv12 = 0.5 * 0.12kg*0.7^2=0.0294 J • Dk=K2-K1=0.47-0.294=0.441 J**Wild Monkey II**A monkey starts pushing a 400 N crate of bananas from rest up an inclined plane whose surface is 300 above the horizontal. The coefficient of kinetic friction of the surface is 0.4. This is one smart monkey since he is simultaneously lifting the box while pushing the box in order to decrease friction. The monkey’s resultant force makes an angle of 200 with the surface of the inclined plane. The monkey’s force is NOT enough to lift the box away from the surface so it stays in contact. After the monkey has moved the box through a distance of 5 m, the box’s velocity is 3 m/s. What is the magnitude of the force that monkey is applying to the box in newtons?**Draw It!**v2=3 m/s 200 F v1=0 5 m f 300**v2=3 m/s**200 F v1=0 5 m f 300 Free body Diagrams F*sin(200) n 200 F*cos(200) 400 cos(300) 300 f 400 sin(300)**Derive Equations**F*sin(200) n 0=n+F*sin(200)-400*cos(300) n= 400*cos(300)-F*sin(200) f=mk*(400*cos(300)-F*sin(200)) F*cos(200) 400 cos(300) f 400 sin(300) Fnet=F*cos(200)+0.4*F*sin(200)-0.4*400*cos(300)-400*sin(300) Fnet=1.07*F-338.5**Use Work Energy**n F*sin(200) Dk=K2-K1 K1=0 m=400/9.8=40.8 kg K2= 0.5*40.8*3^2 K2= 183.6 J Dk=183.6 J F*cos(200) 400 cos(300) f 400 sin(300) Fnet=1.07*F-338.5 Wnet= Fnet * d Wnet= (1.07*F-338.5)*5 m Wnet=5.38*F-1692.8 Dk=Wnet 183.6 = 5.38*F-1692.8 1876.4=5.38*F 348.6N=F**Hints**• 6-22: Make a free body diagram of the forces, find the net work and use the work-energy thm (W=DK) • 6-26: Break weight into components parallel and perpendicular to the surface and then use the work-energy thm. • 6-36: Use the formula found on page 221 (between 6.9 and 6.10) and on the second part, use the work energy thm • 6-60: Part a) use only the work done by gravity to calculate the mass. Part b) use work done by the normal force to calculate the force. Part c) Wtotal=(m*a)*d where d=distance, a=acceleration, m=mass and Wtotal=total work • 6-67: Part a) Plug and chug to find the force at that point. Part b) Plug and chug again. Part c) Use the anti-derivative or integral to integrate the force over x through the interval. • 6-70: Part a)and then use the work-energy thm. Part b) Set up work-energy thm as before and let K2=0 Part c) The total work is zero. • 6-72: Remember the scalar product is F*distance*cosine(angle between them) • 6-80: See Hint on 6-72 and the students work is F*distance*cosine(angle between them). The distance is parallel to the incline. Remember to include the component of weight parallel to the surface. • 6-82: Calculate the total work on both the horizontal block (work done by friction) and sum that with the work done by gravity on the vertical block. W=0.5*(sum of masses)*velocity^2