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A 1.2-kilogram block is connected to a 150 N/m spring on a smooth floor. One end of the spring is connected to a wall. The block is pulled 5 cm to the right and then released. What is the kinetic energy of the block when it is 3 cm from its equilibrium position?


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Transcript
00:00Hi friends, keep up the spirit of pursuing your dreams, even if no one is cheering you on.
00:08Almost similar to the previous series.
00:11But this time, we are asked to calculate the kinetic energy of the block.
00:18Before we go any further, how about we watch this short animation.
00:24A spring block system is undergoing simple harmonic motion.
00:30This block will continue to move to the right and to the left periodically through the equilibrium point.
00:37The block will not move beyond the rightmost line and the leftmost line.
00:41That's because there is no additional external energy and energy dissipation.
00:48If so, the system energy at each point is constant.
00:54Mathematically, E total is equal to the spring energy plus the kinetic energy of the block.
01:02We already know that the total energy of the system is half k A squared.
01:09This A is the maximum displacement of the spring.
01:15And the potential energy of the spring is half k x squared.
01:21From here, the kinetic energy of the block is half k multiplied by A squared minus x squared.
01:30Perhaps you still remember that the spring constant is 150 N per meter.
01:37The maximum displacement of the spring is 0.05 meters.
01:43The displacement of the spring at any time is 0.03 meters.
01:49A little calculation, the kinetic energy is about 0.12 Joules.
01:56This is the kinetic energy of the block.
02:01Happy learning everyone!
02:02Happy learning everyone!
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