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Problem_Set_4_Solutions

# Problem_Set_4_Solutions - 46 We will start by assuming that...

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46. We will start by assuming that the normal force (on the car from the rail) points up. Note that gravity points down, and the y axis is chosen positive upwards. Also, the direction to the center of the circle (the direction of centripetal acceleration) is down. Thus, Newton’s second law leads to 2 . N v F mg m r § · = ¨ ¸ © ¹ (a) When v = 11 m/s, we obtain F N = 3.7 × 10 3 N. (b) N F G points upward. (c) When v = 14 m/s, we obtain F N = –1.3 × 10 3 N, or | F N | = 1.3 × 10 3 N. (d) The fact that this answer is negative means that N F G points opposite to what we had assumed. Thus, the magnitude of N F G is | | N F = G 1.3 kN and its direction is down .

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56. (a) Using the kinematic equation given in Table 2-1, the deceleration of the car is 2 2 2 0 2 0 (35 m/s) 2 (107 m) v v ad a = + ¡ = + which gives 2 5.72 m/s . a = − Thus, the force of friction required to stop by car is 2 3 | | (1400 kg)(5.72 m/s ) 8.0 10 N. f m a = = × (b) The maximum possible static friction is 2 3 ,max (0.50)(1400 kg)(9.80 m/s ) 6.9 10 N. s s f mg μ = = × (c) If 0.40 k μ = , then k k f mg μ = and the deceleration is k a g μ = − . Therefore, the speed of the car when it hits the wall is 2 2 2 0 2 (35 m/s) 2(0.40)(9.8 m/s )(107 m) 20 m/s. v v ad = + = (d) The force required to keep the motion circular is 2 2 4 0 (1400 kg)(35.0 m/s) 1.6 10 N. 107 m r mv F r = = = × (e) Since ,max r s F f > , no circular path is possible.
60. (a) We note that R (the horizontal distance from the bob to the axis of rotation) is the

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