Physics 40 Chapter 8 Homework Q: 12, 13 P: 3, 4, 7, 15, 19, 24, 32, 34, 39, 54, 55, 58, 59, 62, 64

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Corey Leonard
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Physics 40 Chapter 8 Homework Q:, 3 P: 3, 4, 7, 5, 9, 4, 3, 34, 39, 54, 55, 58, 59, 6, 64 Conceptual Questions

In the cases quoted respectiely, this quantity has the numerical alue (a) 5 (b).5 (c).5 (d) 0 (e) 0 () 5.

3 Yes, i it is exerted by an object that is moing in our rame o reerence.

Problems P8.3 U K U K : + = + i i mgh + 0= mg + m g( 3.50) = g + F= : = 3.00g m n+ mg= m 3.00g n= m g = m g =.

1 Physics 40 Chapter 8 Homework Q:, 3 P: 3, 4, 7, 5, 9, 4, 3, 34, 39, 54, 55, 58, 59, 6, 64 Conceptual Questions *Q8. We hae (/)m = μ k mgd so d = /μ k g. The quantity /μ k controls the skidding distance. In the cases quoted respectiely, this quantity has the numerical alue (a) 5 (b).5 (c).5 (d) 0 (e) 0 () 5. In order the ranking is then d > e > = a > b > c. *Q8.3 Yes, i it is exerted by an object that is moing in our rame o reerence. The lat bed o a truck exerts a static riction orce to start a pumpkin moing orward as it slowly starts up. Problems P8.3 U K U K : + = + i i mgh + 0= mg + m g( 3.50) = g + F= : = 3.00g m n+ mg= m 3.00g n= m g = m g =.00mg 3 n = kg 9.80 m s = N downward FIG. P8.3 P8. 4 (a) ( Δ K ) = W = W g = mg Δ h = mg ( ) A B mb ma = m B = 5.94 m s ( 9.80)(.80) Similarly, g C = = 7.67 m s A (b) W g = mg( 3.00 m ) = 47 J A C FIG. P8. 4

2 P8.7 Using conseration o energy or the system o the Earth and the two objects (a) ( 5.00 kg) g( 4.00 m ) = ( 3.00 kg) g( 4.00 m ) + ( ) = 9.6 = 4.43 m s (b) Now we apply conseration o energy or the system o the 3.00 kg object and the Earth during the time interal between the instant when the string goes slack and the instant at which the 3.00 kg object reaches its highest position in its ree all. ( 3.00 ) = mgδ y= 3.00 gδy Δ y =.00 m y max = 4.00 m +Δ y= 5.00 m FIG. P8.7 P8.5 (a) W = mgl cos( θ ) g W = 0.0 kg 9.80 m s 5.00 m cos0 = 68 J g (b) = μ n= μ mgcosθ k k k Δ E = l = lμ mgcosθ int int k k Δ E = 5.00 m cos0.0 = 84 J (c) W Fl F = = = 500 J (d) K Wother Eint WF Wg Eint (e) Δ = Δ = + Δ = Δ K = m m i ( ΔK) 48 J 48 = + (.50) i = + = 5.65 m s m 0.0 FIG. P8.5 P8.9 Ui + Ki +Δ E = U + K : mech mgh h= m + m = μn= μm g mgh μmgh = ( m+ m) ( μ ) m m hg = m + m FIG. P m s.50 m 5.00 kg kg = = 3.74 m s 8.00 kg

P8.4 (a) There is an equilibrium point whereer the graph o potential energy is horizontal: At r=. 5 mm and 3. mm, the equilibrium is stable. At r=. 3 mm, the equilibrium is unstable.

3 P8.4 (a) There is an equilibrium point whereer the graph o potential energy is horizontal: At r=. 5 mm and 3. mm, the equilibrium is stable. At r=. 3 mm, the equilibrium is unstable. A particle moing out toward r approaches neutral equilibrium. (b) The system energy E cannot be less than 5.6 J. The particle is bound i 5.6 J E < J. (c) (d) I the system energy is 3 J, its potential energy must be less than or equal to 3 J. Thus, the particle s position is limited to 0.6 mm r 3.6 mm. K U E. Thus, + = K E U max = min = 3.0 J 5.6 J =.6 J. (e) Kinetic energy is a maximum when the potential energy is a minimum, at r=.5 mm. () 3 J+ W = J. Hence, the binding energy is W = 4 J. P8.3 (a) The distance moed upward in the irst 3.00 s is m s Δ y = t = ( 3.00 s ) =.63 m The motor and the earth s graity do work on the eleator car: mi + Wmotor + mgδ y cos80 = m 4 Wmotor = ( 650 kg )(.75 m s ) 0 + ( 650 kg ) g(.63 m ) =.77 0 J 4 W.77 0 J 3 Also, W = P t so P = = = W = 7.9 hp. t 3.00 s (b) When moing upward at constant speed ( =.75 m s) the applied orce equals the 3 w eight = ( 650 kg)( 9.80 m s ) = N. Thereore, P 3 4 = F = N.75 m s =. 0 W = 4.9 hp P8.34 The useul output energy is 0 Wh 0.60 = mg y y = F Δy i g 0 W s 0.40 J N m Δ y = = 890 N W s J 94 m

4 3 P8.39 (a) x = t+.00t dx = = t Thereore, dt ( 4.00 )( 6.00 ) ( K = m = + t = + t + t ) J d dt (b) a= = ( t).0 m s F= ma= t = 48.0 t N (c) P = F = ( t )( + t ) = ( t + t 3 ) W (d) P W = dt = 48.0t+ 88t dt = 50 J 0 0 P8.54 (a) Between the second and the third picture, Δ Emech =Δ K+Δ U μmgd = m i + kd ( 50.0 N m ) d (.00 kg)( 9.80 m s ) d (.00 kg )( 3.00 m s) = 0 [.45 ±.35 ] N d = = m 50.0 N m (b) Between picture two and picture our, Δ Emech =Δ K+Δ U ( d) = m m i = 3.00 m s.45 N m =.30 m s (.00 kg) (c) For the motion rom picture two to picture ie, Δ E =Δ K+ΔU mech D+ = D ( d) (.00 kg)( 3.00 m s) 9.00 J = ( m) =.08 m FIG. P kg 9.80 m s

5 P8.55 Δ mech E = Δx E E = d i BC mgh kx μ = = mgd kx mgh = μmgd BC BC 0.38 FIG. P8.55 P8.60 (a) Between the second and the third picture, Δ Emech =Δ K +Δ U μmgd= mi + kd ( 50.0 N m ) d (.00 kg)( 9.80 m s ) d (.00 kg)( 3.00 m s ) = 0 [.45±.5 ] N d = = m 50.0 N m (b) Between picture two and picture our, Δ Emech =Δ K +Δ U ( d) = m m i = 3.00 m s.45 N m =.30 m s (.00 kg) (c) For the motion rom picture two to picture ie, Δ E = Δ K +ΔU mech D ( + d) = (.00 kg)( 3.00 m s) 9.00 J D = ( m ) =.08 m kg 9.80 m s FIG. P8.60 P8.58 The geometry reeals D = Lsin θ + Lsin φ, 50.0 m 40.0 m ( sin 50 sin φ ) (a) From takeo to alighting or the Jane-Earth system ( K+ Ug) + Wwind = ( K+ Ug) i mi + mg( Lcosθ) + FD ( ) = 0 + mg( Lcosφ) 50 kg i + 50 kg 9.8 m s 40 m cos 50 0 N 50 m = 50 kg 9.8 m s 40 m cos 8.9 = +, φ = kg i.6 0 J J =.7 0 J 947 J i = = 50 kg 6.5 m s (b) For the swing back

6 mi + mg L + FD + = + mg L ( cosφ) 0 ( cosθ) 30 kg 30 kg 9.8 m s i + 40 m cos N 50 m = 30 kg 9.8 m s 40 m cos50 30 kg J J i + = J 6340 J i = = 30 kg 9.87 m s P8.59 (a) Initial compression o spring: kx = m ( 450 N m )( Δ x) = ( kg)(.0 m s) Thereore, Δ x = m (b) Speed o block at top o track: Δ Emech = Δ x FIG. P8.59 mght + m T mgh B + mb = ( π ) T = ( kg)( 9.80 m s )(.00 m ) + ( kg) ( kg)(.0 m s) 0.50 T 4. T = = 4.0 m s ( 7.00 N )( π )(.00 m ) (c) Does block all o at or beore top o track? Block alls i ac < ( 4.0) T ac = = = 6.8 m s.00 Thereore ac > g and the block stays on the track. g P8.6 (a) Energy is consered in the swing o the pendulum, and the stationary peg does no work. So the ball s speed does not change when the string hits or leaes the peg, and the ball swings equally high on both sides. (b) elatie to the point o suspension, L θ Peg d U i = 0, U = mg d ( L d) From this we ind that mg( d L) + m = 0 FIG. P8.7 Also or centripetal motion,

7 m mg= where = L d. Upon soling, we get 3L d =. 5 P8.64 (a) At the top o the loop the car and riders are in ree all: m Fy = may: mg down = down = g Energy o the car-riders-earth system is consered between release and top o loop: Ki + Ugi = K + Ug : 0+ mgh = m + mg gh = g + g h=.50 (b) Let h now represent the height.5 o the release point. At the bottom o the loop we hae mgh = m b or b = gh F mb y = may: nb mg= ( up) m( gh) nb = mg+ At the top o the loop, mgh = m t + mg = gh 4g F = ma : y y t mt nt mg= m nt = mg+ gh 4g m( gh) nt = 5mg FIG. P8.64 Then the normal orce at the bottom is larger by

8 ( ) ( ) m gh m gh nb nt = mg + + 5mg = 6mg

The negative root tells how high the mass will rebound if it is instantly glued to the spring. We want

The negative root tells how high the mass will rebound if it is instantly glued to the spring. We want 8.38 (a) The mass moves down distance.0 m + x. Choose y = 0 at its lower point. K i + U gi + U si + E = K f + U gf + U sf 0 + mgy i + 0 + 0 = 0 + 0 + kx (.50 kg)9.80 m/s (.0 m + x) = (30 N/m) x 0 = (60