Chapter 5-61. Walker3 5.P.045. [544707] Show Details In a tennis serve, a 0.070 kg ball can be accelerated from rest to 35 m/s over a distance of 0.70 m. Find the magnitude for the average force exerted by the racket on the ball during the serve. 61.3 N[Answer]Newton’s second law: F = m a To find F, we need to find a:For constant acceleration motion: s = (vf2 – vi2 ) /(2*a)Rearrange the equation: a = (vf2 – vi2 )/(2*s) = (352-02)/(2*0.70)=875m/s2Therefore, the average force is: F= m a = 0.070*875=61.3N 2. Walker3 5.P.046. [544509] Show Details A 48.0 kg swimmer with an initial speed of 1.80 m/s decides to coast until she comes to rest. If she slows with constant acceleration and stops after coasting 2.00 m, what was the magnitude of the force exerted on her by the water? 38.9 N[Answer]Similar to problem 1:Acceleration a = (vf2-vi2)/(2*s) = (0-1.802)/(2*2.00) =-0.81m/s2Force F = m a = 48.0 * (-0.81) = -38.9NTherefore, the magnitude of the force is 38.9N3. Walker3 5.P.048. [544489] Show Details An object of mass m = 6.10 kg has an acceleration a = (1.17 m/s2)x + (-0.664 m/s2)y. Three forces act on this object: F1, F2, and F3. Given that F1 = (3.50 N)x and F2 = (-1.55 N)x + (2.05 N)y, find F3.( 5.19 N)x + ( -6.1 N)y[Answer]Newton’s second law: ΣF = m a  F1 + F2 + F3 = m a Therefore, F3 = m a – F1 – F2 F3= 6.10*(1.17 x-0.664 y) - 3.50 x – {-1.55 x + 2.05 y} = 5.19 x -6.1y (N)4. Walker3 5.P.049. [544617] Show Details At the local grocery store, you push a 12.6 kgshopping cart. You stop for a moment to add a bag of dog food to your cart. With a force of 12.0 N, you now accelerate the cart from rest through a distance of 2.49 m in 3.00 s. What was the mass of the dog food? 9.09 kg[Answer]Newton’ second law: F = m aTo find m, we need to know F and aFor constant acceleration motion: xf = xi + vit +1/2at2=0+0+1/2a*3.002=2.49Therefore, a = 2.49*2/3.002 = 0.553m/s2The total mass is, m = F/a = 12.6 + mdog_food = 12.0/0.553Thus, mdog_food= 9.09 kg5. Walker3 5.P.050. [544540] Show Details Biomechanical research has shown that when a85 kg person is running, the force exerted on each foot as it strikes the ground can be as great as 2315 N. (a) What is the ratio of the force exerted on the foot by the ground to the person's body weight? 2.78 (b) If the only forces acting on the person are (i) the force exerted by the ground and (ii) the person's weight, what is the magnitude and direction of the person's acceleration? Magnitude 17.4 m/s2 Direction ---Select--- upward downward upward(c) If the acceleration found in part (b) acts for 11.5 ms, what is the resulting change in the vertical component of the person's velocity? 0.2 m/s [Answer](a) W = m g = 85*9.81 N, the ratio = 2315/(85*9.81) = 2.78(b) ΣF = ma = Fground + W  a =(Fground + W)/m = ( 2315 y + (-85*9.81) y)/85 = 17.4m/s2 y (upward)(c) ΔV = a * Δt = 17.4 y * 11.5e-3 = 0.2 m/s6. Walker3 5.P.051. [544654] Show Details To become airborne, a 2.0 g grasshopper requires a takeoff speed of 2.4 m/s. It acquires this speed by extending its hind legs through a distance of 4.3 cm. (a) What is the average acceleration of the grasshopper during takeoff? 67 m/s2 (b) Find the magnitude of the average net force exerted on the grasshopper by its hind legs during takeoff. 0.134 N (c) If the mass of the grasshopper decreases, does the takeoff acceleration increase, decrease, or stay the same? stays thesame (d) If the mass of the grasshopper decreases, does the required takeoff force increase, decrease, or stay the same? decreases Explain. [Answer](a) average acceleration a = (vf2-vi2)/(2*s)=2.42/(2*4.3e-2) = 67m/s2(b) F = ma = 2.0e-3 * 67 = 0.134N(c) Stay the same: a is not related to mass in (a)(d) Decrease, since F = m a, m decreases  F decreases7. Walker3 5.P.052. [544661] Show Details On an aircraft carrier, a jet can be catapulted from 0 to 145 mi/h in 2.00 s. If the average force exerted by the catapult is 5.60 106 N what is the mass of the jet? 1.73e+05 kg[Answer]Average acceleration = (vf – vi)/ Δt = 145*1609/3600 / 2.00 = 32.4m/s2F = ma  m = F/a = 5.60e6/32.4=1.73e5 kg8. Walker3 5.P.054. [544618] Show Details An apple of mass m = 0.15 kg falls out of a tree from a height h = 3.3 m.(a) What is the magnitude of the force of gravity, mg, acting on the apple? 1.47 N (b) What is the apple's speed, v, just before it lands? 8.05 m/s(c) Calculate mgh and (mv2)/2, including both their numerical values and their units. Express their units in terms of kg, m, and s. 4.85595 kg m^2/s^2 [mgh] 4.85595 kg m^2/s^2 [(mv2)/2] (The relationship between these two quantities will be investigated in Chapter 8.) [Answer](a) mg = 0.15 *9.81 = 1.47N(b) h = vf2/2g  vf = sqrt(2*h*g) = sqrt(2*3.3*9.81) = 8.05m/s(c) mgh = 0.15*9.81*3.3 = 4.86 kg m^2/s^2 1/2mv2 = 1/2*0.15*8.052 = 4.86kg m^2/s^29. Walker3 5.P.056. [544680] Show Details Paleontologists estimate that if a Tyrannosaurus rex were to trip and fall, it would have experienced a force of approximately 256,200 N acting on its torso when it hit the ground. Assume the torso has a mass of 3740 kg. (a) Find the magnitude of the torso's upward acceleration as it comes to rest. (For comparison, humans lose consciousness with an acceleration of about 7g.)58.7 m/s2 (b) Assuming the torso is in free fall for a distance 1.59 m as it falls to the ground, how much time is required for the torso to come to rest once it contacts the ground? 0.0952 s [Answer](a) ΣF = m a = F – mg  a = (F-mg)/m = 256200/3740-9.81 = 58.7m/s2(b) the touch down velocity = sqrt ( 2*s*g) = sqrt (2.0*1.59*9.81) The time for the torso to come to rest = touch down velocity / acceleration = sqrt(2.0*1.59*9.81) / 58.7 = 0.0952 s10. Walker3 5.P.022. [544536] Show Details Two crewmen pull a boat through a lock, as shown in Figure 5-25. One crewman pulls with a force of F1 = 130 N at an angle of = 37° relative to the forward direction of the raft. The second crewman, on the opposite sideof the lock, pulls at an angle of 45°. With what force F2 should the second crewman pull so that the net force of the two crewmen is in the forward direction? 111 NFigure 5-25[Answer]Assume right, up to be positive:Total force in parallel = -F1 * sin 37° + F2 * sin 45°Total force in parallel = 0, therefore; F2 = 111N11. Walker3 5.P.044. [544714] Show Details An ant walks slowly away from the top of abowling ball, as shown in Figure 5-31. If the ant starts to slip when the normal force on