PHY 138 - Chapter 4 Solutions

PHY 138 - Chapter 4 Solutions - Chapter 5 Problem 4.1(a The...

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Unformatted text preview: Chapter 5 Problem 4.1 (a) The net external force acting on this object is ( ) N 12 s m . 2 kg . 6 2 = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = = ma F net . (b) If the same force were applied to a 4.0 kg object, the acceleration would b ( ) ( ) 2 s m . 3 kg 4.0 N 12 = = = m F a net . Problem 4.2 The average force exerted on the ball is ( ) ( ) 2 s m 25 s 0.2 s m 10 kg 50 . = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = = Δ Δ = = contact f avg t v m t v m ma F . Problem 4.5 (a) On the moon, the weight of the bag is ( ) lb 83 . lb 00 . 5 6 1 = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = = = earth earth moon moon earth moon earth moon earth moon W g g W g g mg mg W W . (b) Similarly, on Jupiter the weight is ( ) ( ) lb 2 . 13 lb 00 . 5 64 . 2 = = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = earth earth Jupiter Jupiter W g g W . Problem 4.7 The magnitude of the resistive force is found by applying Newton’s Second Law ( ) ( ) N 6 . 9 s m . 2 kg 20 . N 10 2 = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − = − = − = ma F F F F ma prop drag drag prop . Problem 4.11 (a) The acceleration of the boat is found by applying Newton’s Second Law ( ) ( ) ( ) 2 s m 2 . kg 1000 N 1800 N 2000 = − = − = − = m F F a F F ma drag prop drag prop . (b) The distance the boat will move in 10 s when starting from rest is ( ) m 10 s 10 s m 2 . 2 1 2 1 2 2 2 = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = = at d (c) The speed at the end of 10 s will be ( ) s m 2 s 10 s m 2 . 2 = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = = at v Problem 4.12 (a) The net force is found by resolving the vectors. Setting the vector on the left equal to F 1 , the vector on the right equal to F 2 , and the x ‐ axis perpendicular to the car’s original direction, then ( ) ( ) ( ) ( ) N 346 60 sin N 400 60 sin N 200 60 cos N 400 60 cos N 443 100 sin N 450 100 sin N 1 . 78 100 cos N 450 100 cos 2 2 2 2 1 1 1 1 = ° = ° = =...
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This note was uploaded on 07/09/2008 for the course PHY 138 taught by Professor Meitzler during the Summer '08 term at Sam Houston State University.

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PHY 138 - Chapter 4 Solutions - Chapter 5 Problem 4.1(a The...

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