document.write( "Question 1056858: A 1.60 kg block collides with a horizontal weightless spring of force constant 1.30 N/m. The block compresses the spring 4.00 m from the rest position. Assuming that the coefficient of kinetic friction between the block and the horizontal surface is 0.240 , what was the speed of the block (in meters/second) at the instant of collision?
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Algebra.Com's Answer #671967 by ikleyn(52814) $\"\"$ $\"About$

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\n" ); document.write( "A 1.60 kg block collides with a horizontal weightless spring of force constant 1.30 N/m. The block compresses the spring 4.00 m
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document.write( "Let \"v\" be the initial speed of the block before colliding.\r\n" );
document.write( "Then the initial kinetic energy of the block was .\r\n" );
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document.write( "This kinetic energy was partly spent to overcome the friction force  =   on the distance of d = 4 meters,\r\n" );
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document.write( "and partly (the rest) was transformed to the potential energy of the spring E = ,\r\n" );
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document.write( "where  is the given friction coefficient = 0.240, \"k\" is the given spring of force constant = 1.30 N/m, \"g\" is the gravitation acceleration = 9.81 m/s^2.\r\n" );
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document.write( "So, your energy conservation equation for this problem is\r\n" );
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document.write( " =  + .\r\n" );
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document.write( "Substitute the given data, and you will get the equation to solve\r\n" );
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document.write( " =  + .\r\n" );
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document.write( "Thus the setup is done.\r\n" );
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document.write( "Now solve the equation for \"v\".\r\n" );
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document.write( "It is just arithmetic.\r\n" );
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