2014_SP08_Sol08

2014_SP08_Sol08 - 10.14. Model: Assume an ideal spring that...

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10.14. Model: Assume an ideal spring that obeys Hooke’s law. Visualize: Solve: (a) The spring force on the 2.0 kg mass is sp . F k y = - ∆ Notice that y is negative, so sp F is positive. This force is equal to mg , because the 2.0 kg mass is at rest. We have . k y mg - ∆ = Solving for k : 2 ( / ) (2.0 kg)(9.8 m/s )/( 0.15 m ( 0.10 m)) 392 N/m k mg y = - = - - - - = (b) Again using : k y mg - ∆ = 2 e e / (3.0 kg)(9.8 m/s )/(392 N/m) 0.075 m 0.075 m 0.10 m 0.075 m 0.175 m 17.5 cm y mg k y y y y ∆ = - = - - = - = - = - - = - = - The length of the spring is 17.5 cm when a mass of 3.0 kg is attached to the spring. The position of the end of the spring is negative because it is below the origin, but length must be a positive number. 10.34. Model: This is case of free-fall, so the sum of the kinetic and gravitational potential energy does not change as the cannon ball falls.
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Visualize: The figure shows a before-and-after pictorial representation. To express the gravitational potential energy, we put the origin of our coordinate system on the ground below the fortress. Solve: Using y f = 0 and the equation i gi f gf K U K U + = + we get 2 2 2 2 i i f f i i f 2 2 2 f i i 1 1 2 2 2 2 (80 m/s) 2(9.8 m/s )(10 m) 81.2 m/s mv mgy mv mgy v gy v v v gy + = + + = = + = + = Assess:
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2014_SP08_Sol08 - 10.14. Model: Assume an ideal spring that...

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