a by a factor of = 294 requires 1/T, so to increase 1.4 h 294 = h

Transcription

1 IDENTIFY: If the centripetal acceleration matches g, no contact force is required to support an object on the spinning earth s surface. Calculate the centripetal (radial) acceleration /R using = πr/t to 4π R find arad =. T SET UP: T = 4 h. 6 4 π ( m) 3 EXECUTE: (a) arad =.034 m/s = g. ((4 h)(3600 s/h)) (b) Soling Eq. (3.30) for the period T with arad EVALUATE: arad is proportional to that T be multiplied b a factor of 6 4 π ( m) = g, T = = 5070 s =.4 h m/s a b a factor of = 94 requires h.4 h 94 =. /T, so to increase rad IDENTIFY: Relatie elocit problem in two dimensions. His motion relatie to the earth (time displacement) depends on his elocit relatie to the earth so we must sole for this elocit. (a) SET UP: View the motion from aboe. a The elocit ectors in the problem are: r, M/E the elocit of the man relatie to the earth r, the elocit of the water relatie to the earth r, the elocit of the man relatie to the water The rule for adding these elocities is r r r = + M/E The problem tells us that r has magnitude.0 m/s and direction due south. It also tells us that r has magnitude 4. m/s and direction due east. The ector addition diagram is then as shown in Figure b b This diagram shows the ector addition r r r M/E = + and also has r and r in their specified directions. Note that the ector diagram forms a right triangle. (a) To cross the rier the man must trel 800 m due east relatie to the earth. The man s elocit relatie to the earth is r. M/E But, from the ector addition diagram the eastward component of M/E equals = 4. m/s m Thus t = = = 90 s. 4. m/s (b) The southward component of r M/E equals =.0 m/s. Therefore, in the 90 s it takes him to cross the rier the distance south the man trels relatie to the earth is 0 = t = (.0 m/s)(90 s) = 380 m. EVALUATE: If there were no current he would cross in the same time, (800 m)/(4. m/s) 90 s. = The current carries him downstream but doesn t affect his motion in the perpendicular direction, from bank to bank. 3 Bullet strikes tree, soled in class.

2 4 IDENTIFY: The reaction forces in Newton s third law are alwas between a pair of objects. In Newton s second law all the forces act on a single object. SET UP: Let + be downward. m = w/ g. EXECUTE: The reaction to the upward normal force on the passenger is the downward normal force, also of magnitude 60 N, that the passenger eerts on the floor. The reaction to the passenger s weight is the gritational force that the passenger eerts on the earth, upward and also of magnitude 650 N. F m = a 650 N 60 N gies a =.45 m/s. The passenger s acceleration is 0.45 m /s, (650 N)/(9.80 m/s ) downward. EVALUATE: There is a net downward force on the passenger and the passenger has a downward acceleration. 5 IDENTIFY: Here we appl conseration of energ and use kinematics to find the acceleration. We can use Newton s second law to relate the forces and acceleration. (a) SET UP: First use the information gien about the height of the jump to calculate the speed he has at the instant his feet lee the ground using energ conseration, mgh = /m and find = 4.85m/s. (b) SET UP: Now consider the acceleration phase, from when he starts to jump until when his feet lee the ground. EXECUTE: Calculate the erage acceleration: 4.89 m/s 0 ( a ) = = = 6. m/s t s 0 (c) SET UP: Finall, find the erage upward force that the ground must eert on him to produce this erage upward acceleration. (Don t forget about the downward force of grit.) The forces are sketched. EXECUTE: 890 N m = w/ g = = 90.8 kg 9.80 m/s F = ma F mg = m( a) F = m( g + ( a ) ) F = 90.8 kg(9.80 m/s + 6. m/s ) F = 360 N This is the erage force eerted on him b the ground. But b Newton s 3rd law, the erage force he eerts on the ground is equal and opposite, so is 360 N, downward. EVALUATE: In order for him to accelerate upward, the ground must eert an upward force greater than his weight. 6 The change in the car s kinetic energ is the same as the work done on the car b its brakes, -/m = -Fd. F = N. EVALUATE: A er large force is required to stop such a massie object in such a short distance. 7. IDENTIFY: Appl Newton s second law to calculate a. (a) SET UP: The free-bod diagram for the bucket is sketched.

3 The net force on the bucket is T mg, upward. (b) EXECUTE: F = ma gies T mg = ma T mg a = = = = m 4.80 kg 4.80 kg 75.0 N (4.80 kg)(9.80 m/s ) 75.0 N N 5.8 m/s. EVALUATE: The weight of the bucket is 47.0 N. The upward force eerted b the cord is larger than this, so the bucket accelerates upward. 8. (a) IDENTIFY and SET UP: Find the work done b the positie force. Use the work-energ theorem to find the final kinetic energ, and then K = m gies the final speed. EXECUTE: W = K K, so tot K = Wtot + K K = m = (7.00 kg)(4.00 m/s) = 56.0 J The onl force that does work on the wagon is the 0.0 N force. This force is in the direction of the displacement so the force does positie work, W = Fd = 30.0 J. Then K = Wtot + K = 30.0 J J = 86.0 J. K (86.0 J) K = m ; = = = 4.96 m/s m 7.00 kg r r (b) IDENTIFY: Appl F = ma to the wagon to calculate a. Then use constant acceleration equations to calculate the final speed. The free-bod diagram is gien. SET UP: EXECUTE: F = ma F = ma F 0.0 N.43 m/s a = = = m 7.00 kg EVALUATE: The force in the direction of the motion does positie work and the kinetic energ and speed increase. In part (b), the equialent statement is that the force produces an acceleration in the direction of the elocit and this causes the magnitude of the elocit to increase. 9. (a) IDENTIFY and SET UP: Appl energ conseration to the motion of the potato. Let point be where the potato is released and point be at the lowest point in its motion, as shown in Figure a. K + U + W = K + U other

4 a =.50 m The tension in the string is at all points in the motion perpendicular to the displacement, so W T The onl force that does work on the potato is grit, so W other. EXECUTE: K, Thus U = K. mg = m K m, = U = mg, U = g = (9.80 m/s )(.50 m) = 7.00 m/s EVALUATE: is the same as if the potato fell through.50 m. r r (b) IDENTIFY: Appl F = ma to the potato. The potato moes in an arc of a circle so its acceleration is ar, rad where a = R and is directed toward the center of the circle. Sole for one of rad / the forces, the tension T in the string. SET UP: The free-bod diagram for the potato as it swings through its lowest point is gien in Figure b. The acceleration a r rad is directed in toward the center of the circular path, so at this point it is upward. b EXECUTE: F = ma T mg = ma rad T = m( g + arad ) = m g +, where the radius R for the circular motion is the length L of the string. R It is instructie to use the algebraic epression for from part (a) rather than just putting in the numerical alue: = g = gl, so = gl gl Then T = m g + = m g + = 3 mg; the tension at this point is three times the weight of the L L potato. T = = = 3mg 3(0.00 kg)(9.80 m/s ).94 N EVALUATE: The tension is greater than the weight; the acceleration is upward so the net force must be upward. 0. IDENTIFY: There is no net eternal force on the sstem of the two skaters and the momentum of the sstem is consered. SET UP: Let object A be the skater with mass 70.0 kg and object B be the skater with mass 65.0 kg. Let + be to the right, so A = +.00 m/s and B =.50 m/s. After the collision the two objects are r combined and moe with elocit. Sole for.

5 EXECUTE: P = P. maa + mbb = ( ma + mb). m + m (70.0 kg)(.00 m/s) + (65.0)(.50 m/s) A A B B = = = ma + mb 70.0 kg kg 0.67 m/s. The two skaters moe to the left at 0.67 m/s. EVALUATE: There is a large decrease in kinetic energ. If the duration of the collision is known, the force on each skater is determined b calculating the mass times elocit change diided b the duration.. Weight in Santa Monica s. Mt. Eerest, in class.. IDENTIFY: The sum of the ertical forces on the ingot is zero. ρ = m/ V. The buoant force is B = ρ V g. water obj SET UP: The densit of aluminum is.7 0 kg/m. The densit of water is.00 0 kg/m. m 9.08 kg EXECUTE: (a) T = mg = 89 N so m = 9.08 kg. V = = = m = 3.4 L. ρ.7 0 kg/m (b) When the ingot is totall immersed in the water while suspended, T + B mg. B = ρ V g = =. water obj (.00 0 kg/m )( m )(9.80 m/s ) 3.9 N T = mg B = 89 N 3.9 N = 56 N. EVALUATE: The buoant force is equal to the difference between the apparent weight when the object is submerged in the fluid and the actual grit force on the object. 3. IDENTIFY: The ertical forces on the rock sum to zero. The buoant force equals the weight of liquid displaced b the rock. SET UP: The densit of water is.00 0 kg/m. EXECUTE: The rock displaces a olume of water whose weight is 39. N 8.4 N.8 N. The mass of this much water is thus.0 kg.00 0 kg/m 0.8 N/(9.80 m/s ) =.0 kg and its olume, equal to the rock s olume, is =.0 0 m. The weight of unknown liquid displaced is 39. N 8.6 N.6 N, and its mass is.0 kg/(.0 0 m ) =.9 0 kg/m. 0.6 N/(9.80 m/s ) =.0 kg. The liquid s densit is thus EVALUATE: The densit of the unknown liquid is roughl twice the densit of water.