A mass of 1.35 kg rests on a frictionless floor and is attached to a wall with a 25.0-cm long spring with spring constant 1250 N/m.The mass is pulled back away from the wall to stretch the spring to a length of 34.5 cm.The mass is then released.The speed of the block when the spring returns to its original (unstretched)length is
A)10.5 m/s.
B)87.8 m/s.
C)2.89 m/s.
D)1.32 m/s.
E)25.6 m/s.

Question

Quizplus · Accepted Answer

The total energy of the system is conserved. Initially, the spring has a potential energy of (1/2)kx^2, where x is the initial displacement (0.25 m) and k is the spring constant (1250 N/m). At maximum displacement, the potential energy of the spring is (1/2)kx^2, where x is the final displacement (0.345 m). This energy is converted into kinetic energy when the mass is released.

The potential energy of the spring is (1/2)(1250 N/m)(0.25 m)^2 = 39.1 J.
The final potential energy of the spring is (1/2)(1250 N/m)(0.345 m)^2 = 81.5 J.
The initial kinetic energy of the mass is zero.
Therefore, the final kinetic energy of the mass is (81.5 J - 39.1 J) = 42.4 J.
The velocity of the mass can be found using the formula for kinetic energy:
KE = (1/2)mv^2
42.4 J = (1/2)(1.35 kg)v^2
v = sqrt((2*42.4 J)/(1.35 kg)) = 2.89 m/s.

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