Note: Referred equations are from your textbook.

Transcription

1 Note: Referred equations are from your textbook 70 DENTFY: Use energy methods There are changes in both elastic and gravitational potential energy SET UP: K + U + W K U other + Points and in the motion are sketched in Figure The spring force and gravity are the only forces doing work on the cheese, so W other 0 and U U + U grav el Figure EXECUTE: Cheese released from rest implies K 0 At the maximum height v 0 so K 0 U U,el + U,grav y 0 implies U,grav 0 U kx,el (800 N/m)(05 m) 05 J (Here x refers to the amount the spring is stretched or compressed when the cheese is at position ; it is not the x-coordinate of the cheese in the coordinate system shown in the sketch) U U,el + U,grav U,grav mg y, where y is the height we are solving for U,el 0 since now the spring is no longer compressed Putting all this into K + U + Wother K + U gives U,el U, grav 05 J 05 J y 7 m mg (0 kg)(980 m/s ) EVALUATE: The description in terms of energy is very simple; the elastic potential energy originally stored in the spring is converted into gravitational potential energy of the system 730 DENTFY and SET UP: The friction force is constant during each displacement and Eq(6) can be used to calculate work, but the direction of the friction force can be different for different displacements f μk mg (05)(5 kg)(980 m/s ) 3675 N; direction of f r is opposite to the motion EXECUTE: (a) The path of the book is sketched in Figure a Figure a For the motion from you to Beth the friction force is directed opposite to the displacement W fs (3675 N)(80 m) 94 J s r and

2 For the motion from Beth to Carlos the friction force is again directed opposite to the displacement and W 94 J (b) The path of the book is sketched in Figure b Wtot W + W 94 J 94 J 59 J s (80 m) 3 m Figure b f r is opposite to (c) s r, so W fs (3675 N)(3 m) 4 J Figure c For the motion from you to Kim W fs W (3675 N)(80 m) 94 J Figure d For the motion from Kim to you W fs 94 J The total work for the round trip is 94 J 94 J 59 J (d) EVALUATE: Parts (a) and (b) show that for two different paths between you and Carlos, the work done by friction is different Part (c) shows that when the starting and ending points are the same, the total work is not ero Both these results show that the friction force is nonconservative du ( x) 734 DENTFY: Apply F( x) d(/ x) SET UP: x EXECUTE: ( ) d( Gmm / x) d(/ x) Gmm Fx x Gmm The force on m is in the x x-direction This is toward m, so the force is attractive EVALUATE: By Newton's 3 rd law the force on due to m is Gm m x, in the +x -direction m (toward m ) The gravitational potential energy belongs to the system of the two masses 738DENTFY: Apply Eq(76) du SET UP: is the slope of the U versus x graph du EXECUTE: (a) Considering only forces in the x-direction, Fx and so the force is ero when the slope of the U vs x graph is ero, at points b and d (b) Point b is at a potential minimum; to move it away from b would require an input of energy, so this point is stable (c) Moving away from point d involves a decrease of potential energy, hence an increase in kinetic energy, and the marble tends to move further away, and so d is an unstable point EVALUATE: At point b, Fx is negative when the marble is displaced slightly to the right and Fx is positive when the marble is displaced slightly to the left, the force is a restoring force, and the equilibrium is stable At point d, a small displacement in either direction produces a force directed away from d and the equilibrium is unstable /

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4 Note: Referred equations are from your textbook 8 DENTFY: Apply Eq 89 to relate the change in momentum of the momentum to the components of the average force on it SET UP: Let +x be to the right and +y be upward EXECUTE: (a) J Δ p mv mv (045 kg)( [650 m/s]cos m/s) 54 kg m/s x x x x J y Δ py mvy mv y (045 kg)([650 m/s]sin30 0) 47 kg m/s The horiontal component is 54 kg m/s, to the left and the vertical component is 4 7 kg m/s, upward J x 54 kg m/s J y 47 kg m/s (b) Fav- x 8800 N F 3 av-y 690 N 3 Δ t 75 0 s Δ t 75 0 s The horiontal component is 8800 N, to the left, and the vertical component is 690 N, upward EVALUATE: The ball gains momentum to the left and upward and the force components are in these directions 834 DENTFY: There is no net external force on the system of the two skaters and the momentum of the system is conserved SET UP: Let object A be the skater with mass 700 kg and object B be the skater with mass 650 kg Let +x be to the right, so v Ax + 00 m/s and v Bx 50 m/s After the collision the two objects are r combined and move with velocity v Solve for v x EXECUTE: Px Px mv A Ax+ mv B B x ( ma) vx v mv + mv (700 kg)(00 m/s) + (650)( 50 m/s) A Ax B Bx x ma 700 kg kg The two skaters move to the left at 067 m/s EVALUATE: There is a large decrease in kinetic energy 067 m/s 844 DENTFY and SET UP: Without rounding, the calculation in Example 8 gives v B 0 m/s EXECUTE: The two equations in Example 8 for α and β are and (0500 kg)(400 m/s) (0500 kg)(00 m/s)(cos α) + (0300 kg)( 0 m/s)(cos β ) Eq 0 (0500 kg)(00 m/s)(sin α) (0300 kg)( 0 m/s)sin β Eq Dividing each equation by (0500 kg)(00 m/s) gives and cosα cos β Eq sinα 06 0sinβ Eq cosα Eq 3 gives cos β and cos β cosα cos α 06 0 Eq 4 gives sin β 07454sinα and sin β 05556sin α cos α Adding the two equations and using sin β + cos β gives 778 cosα and cosα 0800 α 369 Then sin β 07454sinα gives β 66 EVALUATE: For these values of α and β, the x component of momentum, the y component of momentum and the kinetic energy are all conserved in the collision 850 DENTFY: Apply Eqs 88, 830 and 83 There is only one component of position and velocity SET UP: m A 00 kg, m B 800 kg M ma 3000 kg Let +x be to the right and let the origin be at the center of mass of the station wagon

5 mx A A + mx B B 0 + (800 kg)(400 m) EXECUTE: (a) xcm 40 m ma 00 kg kg The center of mass is between the two cars, 40 m to the right of the station wagon and 60 m behind the lead car 4 (b) P m v + m v (00 kg)(0 m/s) + (800 kg)(00 m/s) kg m/s x A A B B mv + mv (00 kg)(0 m/s) + (800 kg)(00 m/s) A A, x B B, x (c) vcm, x ma 00 kg kg 4 (d) Px Mvcm- x (3000 kg)(68 m/s) kg m/s EVALUATE: 68 m/s, the same as in part (b) The total momentum can be calculated either as the vector sum of the momenta of the individual objects in the system, or as the total mass of the system times the velocity of the center of mass

6 Note: Referred equations are from your textbook 9 DENTFY: n part (b) apply the equation derived in part (a) SET UP: Let the direction the propeller is rotating be positive ω ω0 EXECUTE: (a) Solving Eq (97) for t gives t Rewriting Eq (9) as α θ θ t( ω + ω t) and substituting for t gives 0 0 ω ω 0 ω + ω0 θ θ0 ω0 ( 0 ) ( 0 ) ( 0 ), + ω ω ω ω ω ω α α α which when rearranged gives Eq (9) (b) α ( ω ω0) ( ( 60 s) ( 0 s) ) 800 /s θ θ0 700 ω0 + ω EVALUATE: We could also use θ θ0 t to calculate t 0500 s Then ω ω + α 0 gives α 800 /s, which agrees with our results in part (b) 930 DENTFY: atan rα, v rω and a v r θ θ ω t 0 av- / t ω0 + ω SET UP: When α is constant, ωav- Let the direction the wheel is rotating be positive atan 00 m s EXECUTE: (a) α 500 s r 000 m v 500 m s (b) At t 300 s, v 500 m s and ω 50 s and at t 0, r 000 m v 500 m s + ( 00 m s )(0 300 s) 800 m s, ω 400 s (c) ω t av- (35 s)(300 s) rev (d) v a r (980 m/s )(000 m) 40 m/s This speed will be reached at time 500 m/s 40 m/s 486 s after t 300 s, or at t 786 s (There are many equivalent ways to do 00 m/s this calculation) 4 4 EVALUATE: At t 0, a rω 30 0 m/s At t 300 s, a 5 0 m/s For a g the wheel must be rotating more slowly than at 300 s so it occurs some time after 300 s 944 DENTFY: SET UP: K EXECUTE: (a) (b) ω Use Table 9 to relate to the mass M of the disk 450 rpm 47 /s For a uniform solid disk, MR and K (050 J) ω (47 /s) 005 kg m (005 kg m ) M 0500 kg R (0300 m) MR EVALUATE: No matter what the shape is, the rotational kinetic energy is proportional to the mass of the object 956 DENTFY: Using the parallel-axis theorem to find the moment of inertia of a thin rod about an axis through its end and perpendicular to the rod SET UP: The center of mass of the rod is at its center, and ML M L M cm cm EXECUTE: p + Md L + M L 3 EVALUATE: is larger when the axis is not at the center of mass