Explanation Consider the given diagram of free body First choose the right angle triangle OB 1 P. B 1 P = O B 1 sin 30 0 B 1 P = 2 sin 30 0 = 1 f t Similarly OP=OB 1 cos30 0 O P = O B 1 cos 30 0 O P = 2 cos 30 0 = 1.732 f t Find the length AP. A P = 2 + 1.732 = 3.732 f t Then consider right angle triangle APB 1 . A B 1 = ( P B 1 ) 2 + ( A P ) 2 A B 1 = ( 1 ) 2 + ( 3.732 ) 2 A B 1 = 3.8636 f t Choose the right-angle triangle OB 2 Q. B 2 Q = O B 2 sin 45 0 B 2 Q = 2 sin 45 0 = 1.4142 f t Substitute above values, we get: O Q = O B 2 cos 45 0 O Q = 2 cos 45 0 = 1.4142 f t Find the length of QA. Q A = 2 − 1.4142 = 0.58578 f t From the right-angle triangle AQB 2 . A B 2 = ( Q B 2 ) 2 + ( A Q ) 2 A B 2 = ( 1.4142 ) 2 + ( 0

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Chapter 13.2, Problem 13.67P

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The spring constant if the velocity of bean bag is given below and by using given information.

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The spring constant if the velocity of bean bag is given below and by using given information.

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