136_PartUniversity Physics Solution

# 136_PartUniversity Physics Solution - 5-8 Chapter 5 EXECUTE...

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5-8 Chapter 5 EXECUTE: (a) y y Fm a = applied to the rocket gives rr gm a −= . 2 2 r r 1720 N (125 kg)(9.80 m/s ) 3.96 m/s 125 kg g a m −− == = . (b) y y a = applied to the power supply gives p sp s nmgma . 22 ps ( ) (1.58 kg)(9.80 m/s 3.96 m/s ) 21.7 N nmga =+ = + = . EVALUATE: The acceleration is constant while the thrust is constant and the normal force is constant while the acceleration is constant. The altitude of 120 m is not used in the calculation. Figure 5.16a, b 5.17. IDENTIFY: Use the kinematic information to find the acceleration of the capsule and the stopping time. Use Newton&s second law to find the force F that the ground exerted on the capsule during the crash. SET UP: Let y + be upward. 311 km/h 86.4 m/s = . The free-body diagram for the capsule is given in Figure 15.17. EXECUTE: 0 0.810 m yy , 0 86.4 m/s y v =− , 0 y v = . 00 2( ) y vv a gives 2 0 2 0 0 ( 86.4 m/s) 4610 m/s 470 2( ) 2( 0.810) m y vv ag = = . (b) y y a = applied to the capsule gives a = and 225 ( ) (210 kg)(9.80 m/s 4610 m/s ) 9.70 10 N 471 . ga w = + = (c) 0 0 2 t + ⎛⎞ ⎜⎟ ⎝⎠ gives 0 2 0 2( ) 2( 0.810 m) 0.0187 s 86.4 m/s 0 t = +− + EVALUATE: The upward force exerted by the ground is much larger than the weight of the capsule and stops the capsule in a short amount of time. After the capsule has come to rest, the ground still exerts a force mg on the capsule, but the large 5 9.00 10 N × force is exerted only for 0.0187 s. Figure 5.17 5.18. IDENTIFY: Apply Newton&s second law to the three sleds taken together as a composite object and to each individual sled. All three sleds have the same horizontal acceleration a . SET UP: The free-body diagram for the three sleds taken as a composite object is given in Figure 5.18a and for each individual sled in Figure 5.18b-d. Let x + be to the right, in the direction of the acceleration. tot 60.0 kg m = . EXECUTE: (a) x x a = for the three sleds as a composite object gives tot Pma = and 2 tot 125 N 2.08 m/s 60.0 kg P a m = .

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Applying Newton&s Laws 5-9 (b) x x Fm a = applied to the 10.0 kg sled gives 10 A PT ma −= and 2 10 125 N (10.0 kg)(2.08 m/s ) 104 N A TPm a =− = = . x x a = applied to the 30.0 kg sled gives 2 30 (30.0 kg)(2.08 m/s ) 62.4 N B Tm a == = . EVALUATE: If we apply x x a = to the 20.0 kg sled and calculate a from A T and B T found in part (b), we get 20 AB TTm a . 2 20 104 N 62.4 N 2.08 m/s 20.0 kg TT a m −− = , which agrees with the value we calculated in part (a).
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136_PartUniversity Physics Solution - 5-8 Chapter 5 EXECUTE...

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