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Unformatted text preview: Chapter 56 1. 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 = (v f 2 – v i 2 ) /(2*a) Rearrange the equation: a = (v f 2 – v i 2 )/(2*s) = (35 20 2 )/(2*0.70)=875m/s 2 Therefore, 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 = (v f 2v i 2 )/(2*s) = (01.80 2 )/(2*2.00) =0.81m/s 2 Force F = m a = 48.0 * (0.81) = 38.9N Therefore, the magnitude of the force is 38.9N 3. Walker3 5.P.048. [544489] Show Details An object of mass m = 6.10 kg has an acceleration a = (1.17 m/s 2 ) x + (0.664 m/s 2 ) y . Three forces act on this object: F 1 , F 2 , and F 3 . Given that F 1 = ( 3.50 N) x and F 2 = (1.55 N) x + (2.05 N) y , find F 3 .( 5.19 N) x + ( 6.1 N) y [Answer] Newton ’s second law: ΣF = m a F 1 + F 2 + F 3 = m a Therefore, F 3 = m a – F 1 – F 2 F 3 = 6.10*(1.17 x0.664 y)  3.50 x – {1.55 x + 2.05 y} = 5.19 x 6.1 y (N) 4. Walker3 5.P.049. [544617] Show Details At the local grocery store, you push a 12.6 kg shopping 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 a To find m, we need to know F and a For constant acceleration motion: x f = x i + v i t +1/2at 2 =0+0+1/2a*3.00 2 =2.49 Therefore, a = 2.49*2/3.00 2 = 0.553m/s 2 The total mass is, m = F/a = 12.6 + m dog_food = 12.0/0.553 Thus, m dog_food = 9.09 kg 5. Walker3 5.P.050. [544540] Show Details Biomechanical research has shown that when a 85 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...
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 Spring '11
 BEAN

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