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Unformatted text preview: 10.38. Model: Model the block as a particle and the springs as ideal springs obeying Hooke's law. There is
no friction, hence the mechanical energy K + Us is conserved. Visualize: Note that xf = xe and xi - xe = Dx. The before-and-after pictorial representations show that we put the origin of the coordinate system at the equilibrium position of the free end of the springs. Solve: The conservation of energy equation Kf + U sf = K i + U si for the single spring is 1 1 1 2 1 2 2 2 mv f + k ( xf - xe ) = mv i + k( x i - x e ) 2 2 2 2 Using the value for vf given in the problem, we get 1 1 1 1 2 2 2 2 mv 0 + 0 J = 0 J + k (Dx ) fi mv 0 = k (Dx ) 2 2 2 2 Conservation of energy for the two-spring case: 1 1 1 1 2 2 2 2 2 mVf + 0 J = 0 J + k ( xi - x e ) + k ( xi - x e ) mVf = k( Dx) 2 2 2 2 Using the result of the single-spring case, 1 2 2 mVf = mv 0 fi Vf = 2v 0 2 Assess: The block separates from the spring at the equilibrium position of the spring. 10.54 10.54. Model: This is a two-part problem. In the first part, we will find the critical velocity for the ball to go over the top of the peg without the string going slack. Using the energy conservation equation, we will then obtain the gravitational potential energy that gets transformed into the critical kinetic energy of the ball, thus determining the angle q. Visualize: We place the origin of our coordinate system on the peg. This choice will provide a reference to measure gravitational potential energy. For q to be minimum, the ball will just go over the top of the peg. Solve: The two forces in the free-body force diagram provide the centripetal acceleration at the top of the circle. Newton's second law at this point is
w +T = mv 2 r where T is the tension in the string. The critical speed vc at which the string goes slack is found when T 0. In this case,
mg = mv 2 C r 1 2 fi vC = gr = gL 3
2 The ball should have kinetic energy at least equal to
mv c =
2 1 2 L^ mg ~ 3 for the ball to go over the top of the peg. We will now use the conservation of mechanical energy equation to get the minimum angle q. The equation for the conservation of energy is
Kf + U gf = K i + U gi fi
Using v f = vc , y f =
1 3 1 2 L 3 mv f + mgy f =
v c , we get
2 2 1 2 mv i + mgy i 2 L, v i = 0, and the above value for 1 2 mg L 3 + mg = mgy i fi yi = L 2 That is, the ball is a vertical distance 1 2 L above the peg's location or a distance of
L^ L 2L - ~= 3 2 6 below the point of suspension of the pendulum, as shown in the figure on the right. Thus, L /6 1 cos q = = fi q = 80.4 L 6 ...
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