ch04 - Chapter 4 The Energy of Particles 4.1 Kinetic Energy...

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Unformatted text preview: Chapter 4 The Energy of Particles 4.1 Kinetic Energy 75 4.1.1 GOAL: Find the work performed by the lift. GIVEN: m = 1100 kg, ¨ y = 0 . 9 m/s 2 , y 2- y 1 = 2 . 0 m DRAW FORMULATE EQUATIONS: The work performed from state 1 to state 2 is given by W = Z y 2 y 1 Ndy (1) The balance of forces on the car is N- mg = m ¨ y ⇒ N = m ( g + ¨ y ) (2) SOLVE: (2) → (1) ⇒ W = m ( g + ¨ y )( y 2- y 1 ) W = (1100 kg)(9 . 81 m/s 2 + 0 . 9 m/s 2 )(2 . 0 m) = 23 , 562 N · m W = 23 , 562 N · m 76 4.1.2 GOAL: Find the speed of the mass when it is 0 . 6 m from the wall GIVEN: m = 0 . 5 kg, k = 40 N/m, unstretched length= L = 0 . 3 m, x 1 = 2 m, x 2 = 0 . 6 m DRAW: ASSUME: The surface is frictionless and the spring is linear. FORMULATE EQUATIONS: We’ll apply work/energy: KE 2 = KE 1 + W 1- 2 KE 1 = 0 ⇒ KE 2 = W 1- 2 SOLVE: W 1- 2 = Z s 2 s 1 F t dt = Z x 2 x 1- ( k ( x- L )) dx W 1- 2 =- k Z x 2 x 1 ( x- L ) dx =- k [ x 2 2- L x ] x 2 x 1 W 1- 2 =- (40 N/m)[( (0 . 6 m) 2 2- (0 . 3 m)(0 . 6 m) )- ( (2 m) 2 2- (0 . 3 m)(2 m) )] = 56 J KE 2 = W 1- 2 = 56 J = 1 2 mv 2 ⇒ v = 14 . 97 m/s * v =- 14 . 97 * ı m/s 77 4.1.3 GOAL: Find speed of arrow after it has moved 1 . 6 feet GIVEN: arrow’s weight and force profile DRAW: FORMULATE EQUATIONS: KE 2 = KE 1 + W 1- 2 SOLVE: m = (20 oz) (16 oz/lb)(32 . 2 ft/s 2 ) = 3 . 88 × 10- 3 slug KE 1 = 0 W 1- 2 = [ 1 . 6 Z 40 e- 3 . 2 x dx ]ft · lb = - 40 3 . 2 e- 3 . 2 x 1 . 6 ft · lb = 12 . 4 ft · lb 1 2 m ˙ x 2 = W 1- 2 1 2 (3 . 88 × 10- 3 slug) ˙ x 2 = 12 . 4 ft · lb ˙ x = 80 ft/s 78 4.1.4 GOAL: Find the speed of a mass after being acted on by a given force. GIVEN: Initial speed, mass, force profile DRAW: FORMULATE EQUATIONS: KE 2 = KE 1 + W 1- 2 SOLVE: KE 1 = 1 2 mv 2 = 1 2 (50 kg)(3 m/s) 2 = 225 N · m W 1- 2 =- 2 Z [- 100- 50 e- 1 . 1 x (2 + x ) ] dx Evaluating with the MATLAB M-file quad yields W 1- 2 =- 113 N · m 1 2 (50 kg) ˙ x 2 = 225 N · m- 113 N · m ˙ x = 2 . 12 m/s 79 4.1.5 GOAL: Plot speed and acceleration of a cyclist going downhill GIVEN: Mass, slope, and drag forces DRAW: * ı * * b 1 cos θ sin θ * b 2- sin θ cos θ FORMULATE EQUATIONS: Force balance:- mg * + F * b 1 + N * b 2 =- m ¨ s * b 1- mg (sin θ * b 1 + cos θ * b 2 ) + (5 N + a ˙ s 2 ) * b 1 + N * b 2 =- m ¨ s * b 1 * b 1 : m ¨ s = mg sin θ- 5 N- a ˙ s 2 (1) * b 2 :- mg cos θ + N = 0 (2) Work/displacement: Z s 2 s 1 Fds = W 1- 2 (3) SOLVE: (1) ⇒ ¨ s = g sin θ- 5 N m- . 04 ˙ s 2 m (4) Using (4) in MATLAB ( θ = 6 ◦ , a = 0 . 04 N · s 2 /m 2 , m = 50 kg) produces the required s, ˙ s data. Using this data along with (4) then lets us solve for ¨ s and produce the following two plots. 80 To calculate the work done by drag forces we employ W drag =- Z 50 m 5 N + a ˙ s 2 ds Using the speed/displacement data already obtained from the numerical simulation lets us calculate the work done and produces a final result of W drag =- 340 N · m 81 4.1.6 GOAL: Determine the work needed to lift a block....
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ch04 - Chapter 4 The Energy of Particles 4.1 Kinetic Energy...

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