146_PartUniversity Physics Solution

146_PartUniversity - 5-18 5.38 Chapter 5 f = r n Apply IDENTIFY SET UP F = ma to the tire n = mg and f = ma 2 v 2 v0 where L is the distance

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5-18 Chapter 5 5.38. IDENTIFY: r f n μ = . Apply m = F a ! ! to the tire. SET UP: nm g = and f ma = . EXECUTE: 22 0 x vv a L = , where L is the distance covered before the wheel&s speed is reduced to half its original speed and 0 /2 = . 2 1 00 4 r 3 8 avv v g Lg Lg Lg == = = . Low pressure, 18.1 m L = and 2 2 3 (3.50 m/s) 0.0259 8 (18.1 m)(9.80 m/s ) = . High pressure, 92.9 m L = and 2 2 3 (3.50 m/s) 0.00505 8 (3.50 m/s) = . EVALUATE: r is inversely proportional to the distance L , so r1 2 r2 1 L L = . 5.39. IDENTIFY: Apply m = F a ! ! to the box. Use the information about sliding to calculate the mass of the box. SET UP: kk f n = , rr f n = and g = . EXECUTE: Without the dolly: g = and k 0 Fn = (0 x a = since speed is constant). 2 k 160 N 34.74 kg (0.47) (9.80 m s ) F m g = With the dolly: the total mass is 34.7 kg 5.3 kg 40.04 kg + = and friction now is rolling friction, . f mg = r Fm g m a −= . 2 r 3.82 m s g a m . EVALUATE: 160 N fm g and 4.36 N g , or, f f = . The rolling friction force is much less than the kinetic friction force. 5.40. IDENTIFY: Apply m = F a ! ! to the truck. For constant speed, 0 a = and horiz r Ff = . SET UP: r f g μμ . Let 21 1.42 mm = and r2 r1 0.81 = . EXECUTE: Since the speed is constant and we are neglecting air resistance, we can ignore the 2.4 m/s, and net F in the horizontal direction must be zero. Therefore r r horiz 200 N fn F = before the weight and pressure changes are made. After the changes, r horiz (0.81 ) (1.42 ) , nF = because the speed is still constant and net 0 F = . We can simply divide the two equations : r horiz r (0.81 )(1.42 ) 200 N μ n = and horiz (0.81) (1.42) (200 N) 230 N F . EVALUATE: The increase in weight increases the normal force and hence the friction force, whereas the decrease in r reduces it. The percentage increase in the weight is larger, so the net effect is an increase in the friction force. 5.41. IDENTIFY: Apply m = F a ! ! to each block. The target variables are the tension T in the cord and the acceleration a of the blocks. Then a can be used in a constant acceleration equation to find the speed of each block. The magnitude of the acceleration is the same for both blocks. SET UP: The system is sketched in Figure 5.41a. For each block take a positive coordinate direction to be the direction of the block&s acceleration. Figure 5.41a
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Applying Newton&s Laws 5-19 block on the table : The free-body is sketched in Figure 5.41b. EXECUTE: y y Fm a = 0 A nmg = A nm g = kk k A f g μμ == Figure 5.41b x x a = k A Tf m a −= k AA Tm g m a μ SET UP: hanging block : The free-body is sketched in Figure 5.41c. EXECUTE: y y a = B B mg T ma B B gm a =− Figure 5.41c (a) Use the second equation in the first k BB mg ma −− = k () ( ) AB B A mm am m g += 2 2 k ( ) (1.30 kg (0.45)(2.25 kg))(9.80 m/s ) 0.7937 m/s 2.25 kg 1.30 kg BA g a = ++ SET UP: Now use the constant acceleration equations to find the final speed. Note that the blocks have the same speeds. 0 0.0300 m, xx 2 0.7937 m/s , x a = 0 0, x v = ?
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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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146_PartUniversity - 5-18 5.38 Chapter 5 f = r n Apply IDENTIFY SET UP F = ma to the tire n = mg and f = ma 2 v 2 v0 where L is the distance

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