SOLUTION: 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 coefficien
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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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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Solution. ("by Physicist")
Let "v" be the initial speed of the block before colliding.
Then the initial kinetic energy of the block was .
This kinetic energy was partly spent to overcome the friction force = on the distance of d = 4 meters,
and partly (the rest) was transformed to the potential energy of the spring E = ,
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.
So, your energy conservation equation for this problem is
= + .
Substitute the given data, and you will get the equation to solve
= + .
Thus the setup is done.
Now solve the equation for "v".
It is just arithmetic.
