141_PartUniversity Physics Solution

141_PartUniversity Physics Solution - Applying Newtons Laws...

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Applying Newton&s Laws 5-13 EVALUATE: The table is level in the direction along its length, since the velocity in that direction is constant. The angle of slope to the right is small, so the acceleration and deflection in that direction are small. 5.26. IDENTIFY: Acceleration and velocity are related by y y dv a dt = . Apply m F =a ! ! to the rocket. SET UP: Let y + be upward. The free-body diagram for the rocket is sketched in Figure 5.26. F ! is the thrust force. EXECUTE: (a) 2 y vA tB t =+ . 2 y aAB t . At 0 t = , 2 1.50 m/s y a = so 2 1.50 m/s A = . Then 2.00 m/s y v = at 1.00 s t = gives 22 2.00 m/s (1.50 m/s )(1.00 s) (1.00 s) B and 3 0.50 m/s B = . (b) At 4.00 s t = , 23 2 1.50 m/s 2(0.50 m/s )(4.00 s) 5.50 m/s y a = . (c) y y Fm a = applied to the rocket gives Tm gm a = and 4 ( ) (2540 kg)(9.80 m/s 5.50 m/s ) 3.89 10 N ag = + . 1.56 Tw = . (d) When 2 1.50 m/s a = , 4 (2540 kg)(9.80 m/s 1.50 m/s ) 2.87 10 N T = × EVALUATE: During the time interval when 2 () vt A t B t applies the magnitude of the acceleration is increasing, and the thrust is increasing. Figure 5.26 5.27. IDENTIFY: Consider the forces in each case. There is the force of gravity and the forces from objects that touch the object in question. SET UP: A surface exerts a normal force perpendicular to the surface, and a friction force, parallel to the surface. EXECUTE: The free-body diagrams are sketched in Figure 5.27a-c. EVALUATE: Friction opposes relative motion between the two surfaces. When one surface is stationary the friction force on the other surface is directed opposite to its motion. Figure 5.27a&c 5.28. IDENTIFY: ss f n μ and kk f n = . The normal force n is determined by applying m F ! ! to the block. Normally, ks μμ . s f is only as large as it needs to be to prevent relative motion between the two surfaces. SET UP: Since the table is horizontal, with only the block present 135 N n = . With the brick on the block, 270 N n = . EXECUTE: (a) The friction is static for 0 P = to 75.0 N P = . The friction is kinetic for 75.0 N P > . (b) The maximum value of s f is s n . From the graph the maximum s f is s 75.0 N f = , so s s max 75.0 N 0.556 135 N f n == = . f n = . From the graph, k 50.0 N f = and k k 50.0 N 0.370 135 N f n = . (c) When the block is moving the friction is kinetic and has the constant value f n = , independent of P . This is why the graph is horizontal for 75.0 N P > . When the block is at rest, s f P = since this prevents relative motion. This is why the graph for 75.0 N P < has slope 1. + (d) s max f and k f would double. The values of f on the vertical axis would double but the shape of the graph would be unchanged.
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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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141_PartUniversity Physics Solution - Applying Newtons Laws...

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