Selected Hw Problems for Ch. 5 and 6

# Selected Hw Problems for Ch. 5 and 6 - 5.4 r m ^ m a =-8.00...

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5.4 22 mm ˆˆ a ( 8.00 )i (6.00 )j ss =− + r 1 F (30.0N)i (16.0N)j =+ r 2 F ( 12.0N)i (8.00N)j + r What must be for the object to have this acceleration? 3 F r Fm a = r r 1 F r 2 F ( 12.0N)i + r 33 3 Fx i y j r ˆ ˆ (30.0 12.0 x )Ni (16.0 8 y )Nj (2.00kg) ( 8.00 (6.00 ⎡⎤ −+ + + + = + ⎢⎥ ⎣⎦ ˆ ˆ (18.0 (24 y )Nj ( 16.00N)i (12.00N)j ++ += + add

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3 18.0N x 16.0N += Which gives 2 equations, 3 24.0N y 12.0N + = 3 x3 4 . 0 N =− 3 y 12.0N = − So, 3 ˆˆ F ( 34.0N)i ( 12.0N)j +− r
5.15 T W = 98N = 58.8 N = 49.0 N = 9.8 N For disk A, Fm a0 == r r WA 1 Tm g T 0 −− = W1 A 2 TT 98N 58.8N m4 k g m g 9.8 s = Similarly for the others, 12 B 2 T T 58.8N 49.0N m1 k g m g 9.8 s = 23 C 2 49.0N 9.8N k g m g 9.8 s = 3 D 2 T 9.8N k g m g 9.8 s =

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5.22 θ F r Fcos θ Fm a = r r ma θ = We’re told that the constant force he pulls with is 2.5 times his body weight. Since he weighs 80 kg this force is (2.5)(80kg)(9.8 m/s 2 )=1960 N. We need the mass of the trains which is m = 700,000N/9.8 m/s 2 = 71,429 kg, so: a m θ = o 2 2 1960Ncos30 m a2 . 3 8 x 1 0 m 71429 s θ == = 22 oo vv2 a ( x x ) =+ Now apply, 2 m v 0 2(2.38x10 )(1m) s 2 2 mm v 2(2.38x10 ) 0.22 ss
5.36 x θ θ mg mgsin θ W F xx Fm a = Wx g s i n m a −θ = 40 kg o 10 θ= a) const. v a x = 0 W g s i n o W 2 m F (40kg)(9.8 )sin10 s = W F6 8 N = b) The skier is skiing down slope with a magnitude of a x = 1 m/s 2 then, W g s i n 0 = g s i n m ( a ) = o x 22 mm F m(a ) m(gsin a ) (40kg)((9.8 sin10 1 ) 28N ss −= θ = = c) The skier is skiing down slope with a magnitude of a x = 2 m/s 2 then, g s i n m ( a ) = o x F m(a ) a ) (40kg)((9.8 sin10 2 ) 12N θ = = Where the – sign means the wind is acting down the slope

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5.50 12 kg 15 kg 20 kg T 2 = 111N T 2 = 222N 21 2 T( mm ) a =+ 4 1234 mmmm ) a = +++ 2 12 T a (m m ) = + Fm a = 2 4 T ) + + + We can write two equations, & Solve for a in the first, Use this in the second, 41 2 1 2 3 4 2 T ( ) ( ) T += + + + 14 24 22 32 42 mT mT mT mT mT mT −=− ++ 2 1 2 3 2 4 2 1 4 m ( TT ) m Tm T −= + + 134 4 2 m m )T m T m (T T ) + +− = 2 (12kg 15kg 20kg)111N 12kg(222N) m (222N 111N) + = 2 m2 3 k g =
5.60 a) To rise w/constant v ( a =0) T T mg yy Fm a = TTm g0 +− = 2T mg = 2 m (95kg)(9.8 ) mg s T 466N 22 == = b) To rise w/constant a y = 1.30 m/s 2 gm a = mm 95kg(1.30 9.8 ) m(a g) ss T 527N + + = c) To rise w/constant v (

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## This homework help was uploaded on 04/22/2008 for the course PHY 2048 taught by Professor Field during the Spring '08 term at University of Florida.

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Selected Hw Problems for Ch. 5 and 6 - 5.4 r m ^ m a =-8.00...

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