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Answer :
1.79 kg block is attached to an ideal spring with a spring constant of 118 Nm/oscillating on a horizontal frictionless surface. When the spring is 24.0 cm shorter than its equilibrium length, the speed of the block is 1.79 m/s.
What is the maximum speed of the block?We can use the concept of energy conservation. The maximum speed is achieved when the spring is at its equilibrium position. At this point, the spring has maximum potential energy and zero kinetic energy, and the block has maximum kinetic energy and zero potential energy.
Since there is no energy loss due to friction, the energy remains constant throughout the motion.Kinetic energy + Potential energy = ConstantEnergy
= 0.5kx² + 0.5mv²Where,
k = 118 Nm/xx
= 24.0 cm
= 0.24 m (the distance from the equilibrium position)m
= 1.79 kgv
= 1.79 m/sWe need to solve for the maximum speed v.Substituting the given values,0.5(118 Nm/m)(0.24 m)² + 0.5(1.79 kg)v² = 0.5(118 Nm/m)(0 m)² + 0.5(1.79 kg)(1.79 m/s)²Simplifying,20.515
v² = 17.5841v
= √(17.5841 / 20.515)
= 1.203 m/sTherefore, the greatest speed of the block is 1.203 m/s (approx).
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