111_PartUniversity Physics Solution

# 111_PartUniversity Physics Solution - Newtons Laws of...

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Newton&s Laws of Motion 4-3 EXECUTE: x x Fm a = gives 2 48.0 N 16.0 kg 3.00 m/s x x F m a == = . EVALUATE: The vertical forces sum to zero and there is no motion in that direction. 4.10. IDENTIFY: Use the information about the motion to find the acceleration and then use x x a = to calculate m . SET UP: Let x + be the direction of the force. 80.0 N x F = . EXECUTE: (a) , , 0 11.0 m xx −= 5.00 s t = 0 0 x v = . 2 1 00 2 x x x xv t a t −= + gives 2 0 22 2( ) 2(11.0 m) 0.880 m/s . (5.00 s) x a t = 2 80.0 N 90.9 kg 0.880 m/s x x F m a = . (b) and 0 x a = x v is constant. After the first 5.0 s, . 2 0 (0.880 m/s )(5.00 s) 4.40 m/s x vv a t =+= = 2 1 2 (4.40 m/s)(5.00 s) 22.0 m xx v t = = . EVALUATE: The mass determines the amount of acceleration produced by a given force. The block moves farther in the second 5.00 s than in the first 5.00 s. 4.11. IDENTIFY and SET UP: Use Newton&s second law in component form (Eq.4.8) to calculate the acceleration produced by the force. Use constant acceleration equations to calculate the effect of the acceleration on the motion. EXECUTE: (a) During this time interval the acceleration is constant and equal to 2 0.250 N 1.562 m/s 0.160 kg x x F a m = We can use the constant acceleration kinematic equations from Chapter 2. 11 0 (1.562 m/s )(2.00 s) , t =+ 2 so the puck is at 3.12 m. x = 2 0 0 (1.562 m/s )(2.00 s) 3.13 m/s. x t + = (b) In the time interval from to 5.00 s the force has been removed so the acceleration is zero. The speed stays constant at The distance the puck travels is At the end of the interval it is at 2.00 s t = 3.12 m/s. x v = (3.12 m/s)(5.00 s 2.00 s) 9.36 m. x t −= = = 0 9.36 m 12.5 m. = In the time interval from to 7.00 s the acceleration is again At the start of this interval and 5.00 s t = 2 1.562 m/s . x a = 0 3.12 m/s x v = 0 12.5 m. x = (3.12 m/s)(2.00 s) (1.562 m/s )(2.00 s) . t = + 2 0 6.24 m 3.12 m 9.36 m. + = Therefore, at the puck is at 7.00 s t = 0 9.36 m 12.5 m 9.36 m 21.9 m. = + = 2 0 3.12 m/s (1.562 m/s )(2.00 s) 6.24 m/s x t + = EVALUATE: The acceleration says the puck gains 1.56 m/s of velocity for every second the force acts. The force acts a total of 4.00 s so the final velocity is (1.56 m/s)(4.0 s) 6.24 m/s. = 4.12. IDENTIFY: Apply m F =a ! ! . Then use a constant acceleration equation to relate the kinematic quantities. SET UP: Let x + be in the direction of the force. EXECUTE: (a) 2 / (140 N)/(32.5 kg) 4.31 m/s . aFm = (b) 2 1 2 x x x t . With 2 1 0 2 0, 215 m x vx a t = . (c) 0 x t . With . 0 2 / 43.0 m/s x vv a t x t EVALUATE: The acceleration connects the motion to the forces. 4.13. IDENTIFY: The force and acceleration are related by Newton&s second law. SET UP: x x a = , where x F is the net force. 4.50 kg m = . EXECUTE: (a) The maximum net force occurs when the acceleration has its maximum value. . This maximum force occurs between 2.0 s and 4.0 s. 2 (4.50 kg)(10.0 m/s ) 45.0 N a = (b) The net force is constant when the acceleration is constant. This is between 2.0 s and 4.0 s. (c) The net force is zero when the acceleration is zero. This is the case at 0 t = and 6.0 s t = . EVALUATE: A graph of x F versus t would have the same shape as the graph of x a versus t .

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4-4 Chapter 4 4.14. IDENTIFY: The force and acceleration are related by Newton&s second law.
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## This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

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111_PartUniversity Physics Solution - Newtons Laws of...

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