Chapter 6: Problems 5, 6, 8, 38, 43, 49 & 53 5. ssm Suppose in Figure 6. that +1.1 1 3 J o work is done by the orce F (magnitude 3. N) in moving the suitcase a distance o 5. m. At what angle θ is the orce oriented with respect to the ground? SSM REASONING AND SOLUTION obtain 3 1 W 1 1.1 1 J Solving Equation 6.1 or the angle θ, we θ cos cos 4.8 Fs (3. N)(5. m) 6. A person pushes a 16.-kg shopping cart at a constant velocity or a distance o. m. She pushes in a direction 9. below the horizontal. A 48.-N rictional orce opposes the motion o the cart. (a) What is the magnitude o the orce that the shopper exerts? Determine the work done by (b) the pushing orce, (c) the rictional orce, and (d) the gravitational orce. 9. F mg +y s +x REASONING The drawing shows three o the orces that act on the cart: F is the pushing orce that the shopper exerts, is the rictional orce that opposes the motion o the cart, and mg is its weight. The displacement s o the cart is also shown. Since the cart moves at a constant velocity along the +x direction, it is in equilibrium. The net orce acting on it in this direction is zero, ΣF x. This relation can be used to ind the magnitude o the pushing orce. The work done by a constant orce is given by Equation 6.1 as W ( F cos θ ) s, where F is the magnitude o the orce, s is the magnitude o the displacement, and θ is the angle between the orce and the displacement. SOLUTION a. The x-component o the net orce is zero, ΣF x, so that (4.9a) The magnitude o the orce that the shopper exerts is 48. N F 54.9 N. cos 9. cos 9. b. The work done by the pushing orce F is ( θ ) ( )( )( ) W F cos s 54.9 N cos 9.. m 16 J (6.1)
Chapter 6: Problems 5, 6, 8, 38, 43, 49 & 53 c. The angle between the rictional orce and the displacement is 18, so the work done by the rictional orce is ( θ ) ( )( )( ) W cos s 48. N cos 18.. m 16 J d. The angle between the weight o the cart and the displacement is 9, so the work done by the weight mg is W mg cos θ s 16. kg 9.8 m/s cos 9. m J ( ) ( )( )( )( ) **8. Multiple-Concept Example 5 reviews many o the concepts that play roles in this problem. An extreme skier, starting rom rest, coasts down a mountain slope that makes an angle o 5. with the horizontal. The coeicient o kinetic riction between her skis and the snow is.. She coasts down a distance o 1.4 m beore coming to the edge o a cli. Without slowing down, she skis o the cli and lands downhill at a point whose vertical distance is 3.5 m below the edge. How ast is she going just beore she lands? REASONING It is useul to divide this problem into two parts. The irst part involves the skier moving on the snow. We can use the work-energy theorem to ind her speed when she comes to the edge o the cli. In the second part she leaves the snow and alls reely toward the ground. We can again employ the work-energy theorem to ind her speed just beore she lands. SOLUTION The drawing at the right shows the three orces that act on the skier as she glides on the snow. The orces are: her weight mg, the normal orce, and the kinetic rictional orce k. Her displacement is labeled as s. The workenergy theorem, Equation 6.3, is 1 1 W mv mv where W is the work done by the net external mg orce that acts on the skier. The work done by each orce is given by Equation 6.1, W ( F cos θ ) s, so the work-energy theorem becomes k 65. s 5. +y Since cos 9, the third term on the let side can be eliminated. The magnitude k o the kinetic rictional orce is given by Equation 4.8 as k µ kfn. The magnitude o the normal orce can be determined by noting that the skier does not leave the surace o the slope, so a y m/s. Thus, we have that ΣF y, so
Chapter 6: Problems 5, 6, 8, 38, 43, 49 & 53 The magnitude o the kinetic rictional orce becomes k µ k µ k mg cos 5. Substituting this result into the work-energy theorem, we ind that. Algebraically eliminating the mass m o the skier rom every term, setting cos 18 1 and v m/s, and solving or the inal speed v gives ( ) v gs cos 65. µ cos 5. k ( )( ) ( ) 9.8 m/s 1.4 m cos 65.. cos 5. 7.1 m/s The drawing at the right shows her displacement s during ree all. Note that the displacement is a vector that starts where she leaves the slope and ends where she touches the ground. The only orce acting on her during the ree all is her weight mg. The work-energy theorem, Equation 6.3, is 3.5 m θ mg s W mv mv 1 1 The work W is that done by her weight, so the work-energy theorem becomes In this expression θ is the angle between her weight (which points vertically downward) and her displacement. Note rom the drawing that s cos θ 3.5 m. Algebraically eliminating the mass m o the skier rom every term in the equation above and solving or the inal speed v gives ( ) v v + g scos θ ( ) ( )( ) 7.1 m/s + 9.8 m/s 3.5 m 1.9 m/s
Chapter 6: Problems 5, 6, 8, 38, 43, 49 & 53 38. The skateboarder in the drawing starts down the let side o the ramp with an initial speed o 5.4 m/s. Neglect nonconservative orces, such as riction and air resistance, and ind the height h o the highest point reached by the skateboarder on the right side o the ramp. REASONING The distance h in the drawing in the text is the dierence between the skateboarder s inal and initial heights (measured, or example, with respect to the ground), or h h h. The dierence in the heights can be determined by using the conservation o mechanical energy. This conservation law is applicable because non-conservative orces are negligible, so the work done by them is zero (W nc J). Thus, the skateboarder s inal total mechanical energy E is equal to his initial total mechanical energy E : (6.9b) Solving Equation 6.9b or h h, we ind that SOLUTION Using the act that v 5.4 m/s and v m/s (since the skateboarder comes to a momentary rest), the distance h is 1 1 1 v ( ) 1( ) v 5.4 m/s m/s h 1.5 m g 9.8 m/s
Chapter 6: Problems 5, 6, 8, 38, 43, 49 & 53 *43. ssm The drawing shows a skateboarder moving at 5.4 m/s along a horizontal section o a track that is slanted upward by 48 above the horizontal at its end, which is.4 m above the ground. When she leaves the track, she ollows the characteristic path o projectile motion. Ignoring riction and air resistance, ind the maximum height H to which she rises above the end o the track. SSM REASONING To ind the maximum height H above the end o the track we will analyze the projectile motion o the skateboarder ater she leaves the track. For this analysis we will use the principle o conservation o mechanical energy, which applies because riction and air resistance are being ignored. In applying this principle to the projectile motion, however, we will need to know the speed o the skateboarder when she leaves the track. Thereore, we will begin by determining this speed, also using the conservation principle in the process. Our approach, then, uses the conservation principle twice. SOLUTION Applying the conservation o mechanical energy in the orm o Equation 6.9b, we have We designate the lat portion o the track as having a height h m and note rom the drawing that its end is at a height o h.4 m above the ground. Solving or the inal speed at the end o the track gives ( ) ( ) ( ) ( ) ( ) v v + g h h 5.4 m/s + 9.8 m/s m.4 m 4.6 m/s This speed now becomes the initial speed v 4.6 m/s or the next application o the conservation principle. At the maximum height o her trajectory she is traveling horizontally with a speed v that equals the horizontal component o her launch velocity. Thus, or the next application o the conservation principle v (4.6 m/s) cos 48º. Applying the conservation o mechanical energy again, we have
Chapter 6: Problems 5, 6, 8, 38, 43, 49 & 53 Recognizing that h.4 m and h.4 m + H and solving or H give (.4 m) (.4 m) mv + mg + H mv + mg 1 1 H ( ) ( ) v v 4.6 m/s 4.6 m/s cos 48 g 9.8 m/s ( ).6 m **49. A skier starts rom rest at the top o a hill. The skier coasts down the hill and up a second hill, as the drawing illustrates. The crest o the second hill is circular, with a radius o r 36 m. Neglect riction and air resistance. What must be the height h o the irst hill so that the skier just loses contact with the snow at the crest o the second hill? REASONING I air resistance is ignored, the only non-conservative orce that acts on the skier is the normal orce exerted on the skier by the snow. Since this orce is always perpendicular to the direction o the displacement, the work done by the normal orce is zero. We can conclude, thereore, that mechanical energy is conserved. Our solution will be based on this act. SOLUTION The conservation o mechanical energy (Equation 6.9b) speciies that 1 1 mv + mgh mv + mgh Since the skier starts rom rest v m/s. Let h deine the zero level or heights, then the inal gravitational potential energy is zero. This gives mgh 1 mv (1) At the crest o the second hill, the two orces that act on the skier are the normal orce and the weight o the skier. The resultant o these two orces provides the necessary centripetal orce to keep the skier moving along the circular arc o the hill. When the skier just loses contact with the snow, the normal orce is zero and the weight o the skier must provide the necessary centripetal orce. mg
Chapter 6: Problems 5, 6, 8, 38, 43, 49 & 53 mv mg so that v gr () r Substituting this expression or v into Equation (1) gives 1 r 36 m mgh mgr or h 18 m 53. Starting rom rest, a 93-kg ireighter slides down a ire pole. The average rictional orce exerted on him by the pole has a magnitude o 81 N, and his speed at the bottom o the pole is 3.4 m/s. How ar did he slide down the pole? REASONING As the ireighter slides down the pole rom a height h to the ground (h m), his potential energy decreases to zero. At the same time, his kinetic energy increases as he speeds up rom rest (v m/s) to a inal speed v,. However, the upward nonconservative orce o kinetic riction k, acting over a downward displacement h, does a negative amount o work on him: ( cos18 ) W h h nc k k (Equation 6.1). This work decreases his total mechanical energy E. Applying the work-energy theorem (Equation 6.8), with h m and v m/s, we obtain ( 1 ) ( 1 ) W mv + mgh mv + mgh (6.8) nc (1) SOLUTION Solving Equation (1) or the height h gives 1 1 mv k or ( k) or mgh h mv h mg mv h The distance h that the ireighter slides down the pole is, thereore, ( mg ) k ( 93 kg)( 3.4 m/s) h ( 93 kg)( 9.8 m/s ) 81 N 5.3 m