A block with mass M = 3.00 kg sits at rest on a horizontal, frictionless table a

Question

A block with mass M = 3.00 kg sits at rest on a horizontal, frictionless table and is attached to a spring, which has a spring constant of k = 800.0 N/ m . The other end of the spring is attached to a rod that is perpendicular to the table and is free to rotate. The distance between the center of the rod and the center of the block is L = 1.20 m. The block is given a push until its velocity is V(initial) = 5.00 m/s in the direction shown in the figure. Assume that the push is brief enough that the length of the spring does not change during the push. The block-spring system is then left undisturbed. The length of the spring will change at first, but after a time the motion of the block becomes uniform circular motion, and the speed V and radius R=L+X does not change. Assume the mass of the spring is negligible figures are beneath questions.
A.) What is the work done on the block-spring system by the force that pushed the block?
B.) Will the spring be stretched or compressed? In other words, is x positive (stretched) or negative (compressed)? Explain your choice. Do not show any calculations for part B.
C.) Will the final speed V of the block be greater than, less than, or same as V(initial)? Explain your answer in words.
D.) Calculate X and V.

Explanation / Answer

a) Work done = change in energy = 1/2*mv^2 = 1/2*3*5^2 = 37.5 J

b) Due to the centrifugal force, the spring will stretch.

c) It'll be less than Vinitial because some of the kinetic energy will go into spring potential energy.

d) Centrifugal force = Spring force

mv^2/(L+x) = -kx

mv^2 = -kx(L+x)

3*v^2 = -800*x*(1.2 + x)

Also, Energy conservation: 1/2*mv^2 + 1/2*kx^2 = 1/2*m*Vinitial^2

1/2*3*v^2 + 1/2*800*x^2 = 37.5

-1/2*800*x*(1.2 + x) + 1/2*800*x^2 = 37.5

Solving this we get x = -0.078 m

So, v = 4.83 m/s

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