F10P1AChap6Sol

# F10P1AChap6Sol - Energy and Energy Transfer SOLUTIONS TO...

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185 Energy and Energy Transfer SOLUTIONS TO PROBLEMS Section 6.2 Work Done by a Constant Force P6.1 (a) W = F ! r cos " = 16.0 N ( ) 2.20 m ( ) cos 25.0 ° = 31.9 J (b), (c) The normal force and the weight are both at 90° to the displacement in any time interval. Both do 0 work. (d) W ! = 31.9 J + 0 + 0 = 31.9 J P6.2 The component of force along the direction of motion is F cos ! = 35.0 N ( ) cos 25.0 ° = 31.7 N . The work done by this force is W = F cos ( ) " r = 31.7 N ( ) 50.0 m ( ) = 1.59 # 10 3 J . P6.4 (a) W = mgh = 3.35 ! 10 " 5 ( ) 9.80 ( ) 100 ( ) J = 3.28 ! 10 " 2 J (b) Since R = mg , W air resistance = ! 3.28 " 10 ! 2 J Section 6.3 The Scalar Product of Two Vectors P6.5 We must first find the angle between the two vectors. It is: = 360 ° " 118 ° " 90.0 ° " 132 ° = 20.0 ° Then r F ! r v = Fv cos = 32.8 N ( ) ( ) cos20.0 ° or r F ! r v = 5.33 N ! m s = 5.33 J s = 5.33 W FIG. P6.5

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186 Energy and Energy Transfer P6.7 (a) W = r F ! " r r = F x x + F y y = 6.00 ( ) 3.00 ( ) N ! m + # 2.00 ( ) 1.00 ( ) N ! m = 16.0 J (b) ! = cos " 1 r F # \$ r r F \$ r % & ' ( ) * = cos " 1 16 6.00 ( ) 2 + " 2.00 ( ) 2 ( ) 3.00 ( ) 2 + 1.00 ( ) 2 ( ) = 36.9 ° P6.9 (a) r A = 3.00 ˆ i ! 2.00 ˆ j r B = 4.00 ˆ i ! 4.00 ˆ j = cos " 1 r A # r B AB = cos " 1 12.0 + 8.00 13.0 ( ) 32.0 ( ) = 11.3 ° (b) r B = 3.00 ˆ i ! 4.00 ˆ j + 2.00 ˆ k r A = ! 2.00 ˆ i + 4.00 ˆ j cos = r A " r B AB = # 6.00 # 16.0 20.0 ( ) 29.0 ( ) = 156 ° (c) r A = ˆ i ! 2.00 ˆ j + 2.00 ˆ k r B = 3.00 ˆ j + 4.00 ˆ k = cos " 1 r A # r B AB \$ % & ' ( ) = cos " 1 " 6.00 + 8.00 9.00 # 25.0
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F10P1AChap6Sol - Energy and Energy Transfer SOLUTIONS TO...

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