Homework Solutions Physics 8A chpt ,38,41,44,48,50,56,59,76

Transcription

1 Homewor Solutons Physcs 8A chpt. 7,8,,,8,5,56,59,76.7. Model: The orce s conservatve, so t has a potental energy. Vsualze: Please reer to Fgure P.7 or the graph o the orce. Solve: The denton o potental energy s U = W( ). In addton, wor s the area under the orce-versus-dsplacement graph. Thus U = U U = (area under the orce curve). Snce U = at x = m, the potental energy at poston x s U( x ) = (area under the orce curve rom to x). From m to m, the area ncreases lnearly rom N m to 6 N m, so U ncreases rom J to 6 J. At x = m, the area s 7 J. Thus U = 7 J at x = m, and U doesn t change ater that snce the orce s then zero. Between m and m, where F changes lnearly, U must have a quadratc dependence on x (.e., the potental energy curve s a parabola). Ths normaton s shown on the potental energy graph below. (b) Mechancal energy s E = K U. From the graph, U = J at x =. m. The netc energy s K = = (. g) (5 m/s) =.5 J. Thus E = 5.5 J. (c) The total energy lne at 5.5 J s shown on the graph above. (d) The turnng pont occurs where the total energy lne crosses the potental energy curve. We can see rom the graph that ths s at approxmately.5 m. For a more accurate value, the potental energy uncton s U = x J. The TE lne crosses at the pont where x = 5.5, whch s x =.56 m..8. Model: Use the relatonshp between orce and potental energy and the wor-netc energy theorem. Vsualze: Please reer to Fgure P.8. We wll nd the slope n the ollowng x regons: cm < x < cm, < x < cm, < x < 5 cm, 5< x < 7 cm,and 7 < x < 8 cm. Solve: (a) F x s the negatve slope o the U-versus-x graph, or example, or m < x < m du J = = N Fx = N dx. m Calculatng the values o F x n ths way, we can draw the orce-versus-poston graph as shown. x (b) Snce W = F dx= area o the Fx-versus-x graph between x and x, the wor done by the orce as the partcle moves x rom x = cm to x = 6 cm s J. x (c) The conservaton o energy equaton s K U = K U. We can see rom the graph that U = J and U = J n movng rom x = cm to x = 6 cm. The nal speed s v = m/s, so J (. g)(. m/s) = J (. g) v v =. m/s

2 .. Model: Model the elevator as a partcle. Vsualze: Solve: (a) The wor done by gravty on the elevator s W U mgy mgy mg y y g = g = = ( ) = ( g)(9.8 m/s )( m) = 9.8 J (b) The wor done by the tenson n the cable on the elevator s To nd T we wrte Newton s second law or the elevator: (c) The wor-netc energy theorem s W = T( y)cos = T( y y ) = T( m) T y = G = y = G y = ( y) = ( g)(9.8 m/s. m/s ) F T F ma T F ma m g a 5.8 N WT (.8 N)( m).8 J = = = Wnet = Wg WT = K = K K = K K = Wg WT K = = (d) K =. J = ( g) v v =.5 m/s 5 ( 9.8 J) (.8 J) ( g)( m/s). J.. Model: Model Sam strapped wth ss as a partcle, and apply the law o conservaton o energy. Vsualze: Solve: (a) The conservaton o energy equaton s The snow s rctonless, so Thus, E th = J. K Ug Eth = K Ug Wext However, the wnd s an external orce dong wor on Sam as he moves down the hll.

3 W = W = ( K U ) ( K U ) ext wnd g g = mgy mgy = J ( J mgy) = mgy W v = gy m wnd We compute the wor done by the wnd as ollows: r r Wwnd = Fwnd = Fwnd rcos6 = ( N)(6 m)cos6 = 7, J where we have used r = h/sn = 6 m. Now we can compute ( 7, J) v = = 75 g (9.8 m /s )(5 m) 5.7 m /s (b) We wll use a tlted coordnate system, wth the x-axs parallel to the slope. Newton s second law or Sam s ( Fnet ) x FG sn Fwnd cos mgsn Fwnd cos ax = a= = = m m m (75 g)(9.8 m/s )sn ( N)cos = =.86 m/s 75 g Now we can use constant-acceleraton nematcs as ollows: v = v a( x x ) = ax v = ax = (.86 m/s )(6 m) = 5.7 m/s.8. Model: Identy the truc and the loose gravel as the system. We need the gravel nsde the system because rcton ncreases the temperature o the truc and the gravel. We wll also use the model o netc rcton and the conservaton o energy equaton. Vsualze: We place the orgn o our coordnate system at the base o the ramp n such a way that the x-axs s along the ramp and the y- axs s vertcal so that we can calculate potental energy. The ree-body dagram o orces on the truc s shown. Solve: The conservaton o energy equaton s K U g E. th = K U W g ext In the present case, W ext = J, v x = m/s, U g = J, v x = 5 m/s. The thermal energy created by rcton s E = ( )( x x ) = ( µ n)( x x ) = µ ( mgcos6. )( x x ) Thus, the energy conservaton equaton smples to th = x x = x x (.)(5, g)(9.8 m/s )(cos6. )( ) (58,78 J/m)( ) J mgy (58,78 J/m)( x x) = x J J (5, g)(9.8 m/s )[( x x )sn 6. ] (58,78 J/m)( x x ) = (5, g)(5 m/s) ( x x ) = m

4 .5. Model: Model the two blocs as partcles. The two blocs mae our system. Vsualze: We place the orgn o our coordnate system at the locaton o the. g bloc. Solve: (a) The conservaton o energy equaton s K U E = K U W Usng Eth = J and W ext = J we get g th g ext. m( v) m( v) mg( y) = m( v) m( v) mg( y) Notng that ( v) = ( v) = v and ( v) = ( v) = m/s, ths becomes mg ( y y) (. g)(9.8 m/s )(.5 m) ( m m ) v = mg ( y y ) v = = =. m/s m m (. g. g) (b) We wll use the same energy conservaton equaton. However, ths tme E = m g x = = th µ ( )( ) (.5)(. g)(9.8 m/s )(.5 m) 6.65 J The energy conservaton equaton s now mgy 6.65 J = m ( v) m( v) mgy J ( m m) v 6.65 J = mg( y y) v = [ mg( y y) 6.65 J] m m = 5. g [(. g)(9.8 m/s )(.5 m) 6.65 J]. m/s =.56. Model: Assume an deal sprng, so Hooe s law s obeyed. Treat the physcs student as a partcle and apply the law o conservaton o energy. Our system s comprsed o the sprng, the student, and the ground. We also use the model o netc rcton. Vsualze: We place the orgn o the coordnate system on the ground drectly below the end o the compressed sprng that s n contact wth the student.

5 Solve: (a) The energy conservaton equaton s Snce y = y = m, x = x, e v = m/s, K U U E = K U U W g s th g s ext mgy ( x xe) J = mgy ( x x) J = 8, N/m, g, m = and ( x x) =.5 m, = ( x x ) v = ( x x ) =. m /s m (b) Frcton creates thermal energy. Applyng the conservaton o energy equaton once agan: K U U E = K U U W g s th g s ext mgy J s = J mgy ( x x) J Wth v = m/s and y = ( s)sn, the above equaton s smpled to mg( s)sn n s mgy ( x x ) µ = From the ree-body dagram or the physcs student, we see that n = F G cos = mg cos. Thus, the conservaton o energy equaton gves s( mg sn µ mg cos ) = mgy ( x x ) Usng m = g, = 8, N/m, ( x x) =.5 m, y = m, and µ =.5, we get mgy ( x x ) s = =. m mg (sn µ cos ).59. Model: A sprong that obeys the orce law Fx = q( x xe), where q s the sprong constant and x e s the equlbrum poston. Vsualze: We place the orgn o the coordnate system on the ree end o the sprong, that s, xe = x = m. Solve: (a) The unts o q are N/m. (b) A cubc curve rses more steeply than a parabola. The orce ncreases by a actor o 8 every tme x ncreases by a actor o. du x qx (c) Snce Fx =, we have U ( x) = Fxdx ( qx ) dx. dx = = 5

6 (d) Applyng the energy conservaton equaton to the ball and sprong system: K U = K U qx J = J qx (, N/m ) (. m) v = = = m/s m (. g).76. Solve: (a) The graph s a hyperbola. (b) The separaton or zero potental energy s r =, snce Gmm U = J as r r Ths maes sense because two masses don t nteract at all they are nntely ar apart. (c) Due to the absence o nonconservatve orces n our system o two partcles, the mechancal energy s conserved. The equatons o energy and momentum conservaton are Gmm Gmm K Ug = K Ug = r r = Gmm r r m p = p = g m /s v = v m Substtutng ths expresson or v nto the energy equaton, we get m Gm m v = Gmm v = m r r ( m/ m) r r 8 Wth G = 6.67 N m(g), r= R R= 8 m, r=. m, m= 8. g, and m=. g, the above equaton can be smpled to yeld 5 m 8. g 5 5 v =.7 m/s, and v = v = (.7 m/s) = 6.89 m/s m. g 6

7 5 5 The speed o the heaver star s.7 m/s. That o the lghter star s 6.9 m/s. 7