Ch. 4: FOC 9, 13, 16, 18. Problems 20, 24, 38, 48, 77, 83 & 115;

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1 WEEK-3 Recitation PHYS 3 eb 4, 09 Ch. 4: OC 9, 3,, 8. Pobles 0, 4, 38, 48, 77, 83 & 5; Ch. 4: OC Questions 9, 3,, (e) Newton s law of gavitation gives the answe diectl. ccoding to this law the weight is diectl popotional to the ass of the planet, so twice the ass eans twice the weight. Howeve, this law also indicates that the weight is invesel popotional to the squae of the planet s adius, so thee ties the adius eans one ninth the weight. Togethe, these two factos ean that the weight on the planet is /9 o 0. ties ou eath-weight. 3. (b) ccoding to Newton s thid law, the pushe and the wall eet foces of equal agnitude but opposite diections on each othe. The noal foce is the coponent of the wall s foce that is pependicula to the wall. Thus, it has the sae agnitude as the coponent of the pushe s foce that is pependicula to the wall. s a esult, the noal foces ae anked in the sae ode as the pependicula coponents of the pushe s foces. The sallest pependicula coponent is in B, and the lagest is in C.. (c) The agnitude of the kinetic fictional foce is popotional to the agnitude of the noal foce. The noal foce is sallest in B, because the vetical coponent of copensates fo pat of the block s weight. In contast, the noal foce is geatest in C, because the vetical coponent of adds to the weight of the block. 8. (e) In B the tension T is the sallest, because thee ope segents suppot the weight W of the block, with the esult that 3T W, o T W/3. In the tension is the geatest, because onl one ope segent suppots the weight of the block, with the esult that T W.

2 Ch. 4: Pobles 0, 4, 38, 48, 77, 83 & 5 0. RESONING The gavitational foce acting on each object is specified b Newton s law of univesal gavitation. The acceleation of each object when eleased can be deteined with the aid of Newton s second law. We ecognize that the gavitational foce is the onl foce acting on eithe object, so that it is the net foce to use when appling the second law. a SOLUTION a. The agnitude of the gavitational foce eeted on the ock b the eath is given b Equation 4.3 as eath ock ock eath (.7 0 N / kg )( kg)( 5.0 kg) (.38 0 ) 49 N The agnitude of the gavitational foce eeted on the pebble b the eath is eath pebble pebble eath ( )( N / kg kg)( kg) (.38 0 ).9 0 N b. ccoding to the second law, the agnitude of the acceleation of the ock is equal to the gavitational foce eeted on the ock divided b its ass. a ock eath ock ock eath 4 (.7 0 N / kg )( kg) (.38 0 ) 9.80 /s ccoding to the second law, the agnitude of the acceleation of the pebble is equal to the gavitational foce eeted on the pebble divided b its ass. pebble eath pebble pebble eath 4 (.7 0 N / kg )( kg) (.38 0 ) 9.80 /s 3

3 4. RESONING Newton s law of gavitation shows how the weight W of an object of ass is elated to the ass M and adius of the planet on which the object is located: W GM/. In this epession G is the univesal gavitational constant. Using the law of gavitation, we can epess the weight of the object on each planet, set the two weights equal, and obtain the desied atio. SOLUTION ccoding to Newton s law of gavitation, we have The ass of the object, being an intinsic popet, is the sae on both planets and can be eliinated algebaicall fo this equation. The univesal gavitational constant can likewise be eliinated algebaicall. s a esult, we find that B o M B B B M M M B M M B 38. RESONING In each case the object is in equilibiu. ccoding to Equation 4.9b, Σ 0, the net foce acting in the (vetical) diection ust be zeo. The net foce is coposed of the weight of the object(s) and the noal foce eeted on the. SOLUTION a. Thee ae thee vetical foces acting on the cate: an upwad noal foce + N that the floo eets, the weight g of the cate, and the weight g of the peson standing on the cate. Since the weights act downwad, the ae assigned negative nubes. Setting the su of these foces equal to zeo gives The agnitude of the noal foce is N g + g (35 kg + 5 kg)(9.80 /s ) 980 N

4 b. Thee ae onl two vetical foces acting on the peson: an upwad noal foce + N that the cate eets and the weight g of the peson. Setting the su of these foces equal to zeo gives The agnitude of the noal foce is N g (5 kg)(9.80 /s ) 40 N 48. RESONING ND SOLUTION The deceleation poduced b the fictional foce is fk µ kg a µ k g The speed of the autoobile afte.30 s have elapsed is given b Equation.4 as ( ) ( )( µ )( ) v v0 + at v0 + kg t. /s /s.30 s.9 /s 77. RESONING The onl hoizontal foce acting on the boat and taile is the tension in the hitch; theefoe, it is the net foce. ccoding to Newton s second law, the tension (o the net foce) equals the ass ties the acceleation. The ass is known, and the acceleation can be found b appling an appopiate equation of kineatics fo Chapte 3. SOLUTION ssue that the boat and taile ae oving in the + diection. Newton s second law is Σ a (see Equation 4.a), whee the net foce is just the tension +T in the hitch, so Σ T. Thus, T a () Since the initial and final velocities, v 0 and v, and the tie t ae known, we a use Equation 3.3a fo the equations of kineatics to elate these vaiables to the acceleation: v v + a t (3.3a) 0 Solving Equation (3.3a) fo a and substituting the esult into Equation (), we find that v v0 /s 0 /s T a ( 40 kg) 0 N t 8 s

5 83. SSM RESONING ND SOLUTION The sste is shown in the dawing. We will let.0 kg, and 45.0 kg. Then, will ove upwad, and will ove downwad. Thee ae two foces that act on each object; the ae the tension T in the cod and the weight g of the object. The foces ae shown in the fee-bod diagas at the fa ight. We will take upwad as the positive diection. If the acceleation of is a, then the acceleation of ust be a. o Newton's second law, we have fo that T g T g and fo T g a () T g a () a. Eliinating T between these two equations, we obtain 45.0 kg.0 kg a g (9.80 /s ) 3.5 /s kg +.0 kg b. Eliinating a between Equations () and (), we find (.0 kg)(45.0 kg) T g (9.80 /s ) 8 N +.0 kg kg 5. SSM RESONING Let us assue that the skate is oving hoizontall along the + ais. The tie t it takes fo the skate to educe he velocit to v +.8 /s fo v /s can be obtained fo one of the equations of kineatics: v v + a t (3.3a) 0 The initial and final velocities ae known, but the acceleation is not. We can obtain the acceleation fo Newton s second law ( Σ a, Equation 4.a) in the following anne. The kinetic fictional foce is the onl hoizontal foce that acts on the skate, and, since it is a esistive foce, it acts opposite to the diection of the otion. Thus, the net foce in the diection is Σ fk, whee f k is the agnitude of the kinetic fictional foce. Theefoe, the acceleation of the skate is a Σ / f k /.

6 The agnitude of the fictional foce is f k µ k N (Equation 4.8), whee µ k is the coefficient of kinetic fiction between the ice and the skate blades and N is the agnitude of the noal foce. Thee ae two vetical foces acting on the skate: the upwad-acting noal foce N and the downwad pull of gavit (he weight) g. Since the skate has no vetical acceleation, Newton's second law in the vetical diection gives (taking upwad as the positive diection) Σ N g 0. Theefoe, the agnitude of the noal foce is N g and the agnitude of the acceleation is a fk µ kn µ k g µ k g SOLUTION Solving the equation v v0 + at fo the tie and substituting the epession above fo the acceleation ields v v0 v v0.8 /s.3 /s t 4.4 s a µ g ( )( /s ) k

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