A bead is released with an initial speed of 2.0 m/s. It slides without friction

Question

A bead is released with an initial speed of 2.0 m/s. It slides without friction down a wire and makes a loop-the-loop as shown in the diagram. What is its speed at point A, the bottom of the circular loop, and at point B, the top of the circular loop? A 2.0-kg block on a horizontal, frictionless surface is tied to one end of a massless spring with spring constant 250 N/m as shown in the figure. The block is pulled aside by a force P until the spring has been stretched 20 cm, held at rest and released, a. Find the speed of the block as it passes through its equilibrium position.

Explanation / Answer

A) The initial velocity of the bead, u = 2 m/s The height from where it is dropped, h = 10 m So the total energy at the releasing point of bead is equal to the total energy at point A       mgh + 0.5mv^2 = 0.5mv1^2          gh + 0.5v^2 = 0.5v1^2          9.8(10) + 0.5(2)^2 = 0.5v1^2 From the above we have, v1 = 14.14 m/s Now the total energy at A is equal to the total energy at B       0.5mv1^2 = mgh' + 0.5mv2^2         0.5v1^2 = g(2r) + 0.5v2^2    (r = radius of the circular path = 4 m)         0.5(14.14)^2 - (9.8)(8) = 0.5v2^2 From the above we have, v2 = 6.57 m/s b) The Spring constant, k = 250 Nm The mass, m = 2 kg The expansion in the spring, x = 20 cm = 0.2 m We have a formula 0.5mv^2 = 0.5kx^2 0.5(2)v^2 = 0.5(250)(0.2)^2       From the above we have, v = 2.24 m/s So the velocity of the mass, v = 2.24 m/s        

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