Note: Referred equations are from your textbook.

Transcription

1 Note: Referred equations are from your textboo 50 IDENTIFY: pply Newton s first law to the car SET UP: Use x and y coordinates that are parallel and perpendicular to the ramp EXECUTE: (a) The free-body diagram for the car is given in Figure The vertical weight w and the tension T in the cable have each been replaced by their x and y components (b) F x = 0 gives Tcos30 wsin 50 = 0 and sin 50 sin 50 T = w = (30 g)(980 m/s ) = 5460 N cos30 cos30 (c) F y = 0 gives n + Tsin30 wcos50 = 0 and n= wcos50 - Tsin30 = (30 g)( 980 m/s )cos50 (5460 N)sin30 = 70 N EVLUTE: We could also use coordinates that are horizontal and vertical and would obtain the same values of n and T Figure 58 IDENTIFY: pply Newton s second law to the three sleds taen together as a composite object and to each individual sled ll three sleds have the same horizontal acceleration a SET UP: The free-body diagram for the three sleds taen as a composite object is given in Figure 58a and for each individual sled in Figure b-d Let + x be to the right, in the direction of the acceleration m tot = 600 g EXECUTE: (a) Fx = max for the three sleds as a composite object gives P= mtota and P 5 N 08 m/s a = = = mtot 600 g (b) Fx = max applied to the 00 g sled gives P T = m0aand T = P m0a = 5 N (00 g)(08 m/s ) = 04 N Fx = max applied to the 300 g sled gives T = m30 a = (300 g)(08 m/s ) = 64 N EVLUTE: If we apply Fx = max to the 00 g sled and calculate a from T and T found in part T T 04 N 64 N (b), we get T T = m0a a = = = 08 m/s, which agrees with the value we m0 00 g calculated in part (a)

2 Figure a d 536 IDENTIFY: Constant speed means zero acceleration for each bloc If the bloc is moving the friction r r force the tabletop exerts on it is inetic friction pply F = ma to each bloc SET UP: The free-body diagrams and choice of coordinates for each bloc are given by Figure 3 m = 459 g and m = 55 g EXECUTE: (a) Fy = may with a y = 0 applied to bloc gives mg T= 0 and T = 50 N Fx = max with a x = 0 applied to bloc gives T f = 0 and f = 50 N n = mg = 450 N and f 50 N μ = = = 0556 n 450 N (b) Now let be bloc plus the cat, so m = 98 g n = 900 N and f = μ n= (0556)(900 N) = 500 N Fx = max for gives T f = m a x Fy = may for bloc gives mg T= ma y ax for equals a y for, so adding the two equations gives mg f = ( m + m) a and mg f 50 N 500 N y ay = = = 3 m/s The acceleration is m + m 98 g + 55 g upward and bloc slows down EVLUTE: The equation mg f = ( m + m) ay has a simple interpretation If both blocs are considered together then there are two external forces: mgthat acts to move the system one way and f that acts oppositely The net force of mg f must accelerate a total mass of m + m Figure 3 4π R rad 555 IDENTIFY: The acceleration due to circular motion is a = T SET UP: R = 800 m /T is the number of revolutions per second EXECUTE: (a) Setting a = g and solving for the period T gives rad R 400 m T = π = π = 40 s, g 980 m s

3 so the number of revolutions per minute is (60 s min) (40 s) = 5 rev min (b) The lower acceleration corresponds to a longer period, and hence a lower rotation rate, by a factor of the square root of the ratio of the accelerations, T = (5 rev min) = 09 rev min EVLUTE: In part (a) the tangential speed of a point at the rim is given by a v= Ra = Rg = 66 m/s ; the space station is rotating rapidly rad rad v =, so R

4 Note: Referred equations are from your textboo 6 IDENTIFY: In each case the forces are constant and the displacement is along a straight line, so W = Fscosφ SET UP: In part (a), when the cable pulls horizontally φ = 0 and when it pulls at 350 above the horizontal φ = 350 In part (b), if the cable pulls horizontally φ = 80 If the cable pulls on the car at 350 above the horizontal it pulls on the truc at 350 below the horizontal and φ = 450 For the gravity force φ = 90, since the force is vertical and the displacement is horizontal 3 6 EXECUTE: (a) When the cable is horizontal, W = (850 N)(500 0 m)cos0 = 45 0 J When 3 6 the cable is 350 above the horizontal, W = (850 N)(500 0 m)cos350 = J 6 6 (b) cos80 = cos0 and c os450 = cos350, so the answers are 46 0 J and J (c) Since cosφ = cos90 = 0, W = 0 in both cases EVLUTE: If the car and truc are taen together as the system, the tension in the cable does no net wor 68 IDENTIFY: pply Eq(65) SET UP: iˆ iˆ= ˆj ˆj = and iˆ ˆj = ˆj iˆ= 0 r r EXECUTE: The wor you do is F s = ((30 N) iˆ (40 N) ˆj) (( 90 m) iˆ (30 m) ˆj) r r F s = (30 N)( 90 m) + ( 40 N)( 30 m) = 70 N m + 0 N m = 50 J EVLUTE: The x-component of F r does negative wor and the y-component of F r does positive wor The total 6 IDENTIFY and SET UP: Use Eq(66) to calculate the wor done by the foot on the ball Then use Eq(6) to find the distance over which this force acts EXECUTE: Wtot = K K K = mv = (040 g)(00 m/s) = 084 J K = mv = (040 g)(600 m/s) = 756 J Wtot = K K = 756 J 084 J = 67 J The 400 N force is the only force doing wor on the ball, so it must do 67 J of wor WF = ( Fcos φ) s gives that W 67 J s = 068 m F cos φ = (400 N)(cos0) = EVLUTE: The force is in the direction of the motion so positive wor is done and this is consistent with an increase in inetic energy 68 IDENTIFY: The wor that must be done to move the end of a spring from x to x is The force required to hold the end of the spring at displacement x is Fx = x W = x x SET UP: When the spring is at its unstretched length, x = 0 When the spring is stretched, x > 0, and when the spring is compressed, x < 0 W (0 J) 4 EXECUTE: (a) x = 0 and W = x = = = 67 0 N/m x (00300 m) 4 (b) F = x= (67 0 N/m)(00300 m) = 80 N x 4 (c) x =, x = m W = (67 0 N/m)( m) = 4 J Fx 0 4 = x = (67 0 N/m)(00400 m) = 070 N EVLUTE: When a spring, initially unstretched, is either compressed or stretched, positive wor is done by the force that moves the end of the spring

5 646 IDENTIFY: The thermal energy is produced as a result of the force of friction, F = μ mg The average thermal power is thus the average rate of wor done by friction or SET UP: av v v + v 800 m/s + 0 = = = 400 m/s P = Fv av EXECUTE: P= Fv ( )( )( ) ( ) av = g 980 m/s 400 m/s = 57 W EVLUTE: The power could also be determined as the rate of change of inetic energy, Δ K t, where the time is calculated from vf = vi + at and a is calculated from a force balance, F = ma =μ mg

6 Note: Referred equations are from your textboo 74 IDENTIFY: Only gravity does wor on him from the point where he has just left the board until just before he enters the water, so Eq(74) applies SET UP: Let point be just after he leaves the board and point be just before he enters the water + y is upward and y = 0 at the water EXECUTE: (a) K = 0 y = 0 y = 35 m K+ Ugrav, = K + Ugrav, gives Ugrav, = K and mgy = mv v = = = gy (980 m/s )(35 m) 798 m/s (b) v = 50 m/s, y = 0, y = 35 m K+ Ugrav, = K and v = v + gy = (50 m/s) + (980 m/s )(35 m) = 836 m/s mv + mgy = mv (c) v = 5 m/s and v = 836 m/s, the same as in part (b) EVLUTE: Kinetic energy depends only on the speed, not on the direction of the velocity