Question 1056858
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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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<U>Solution</U>. ("by Physicist")


<pre>
Let "v" be the initial speed of the block before colliding.
Then the initial kinetic energy of the block was {{{(mv^2)/2}}}.

This kinetic energy was partly spent to overcome the friction force {{{F[friction]}}} = {{{mu*mg}}}  on the distance of d = 4 meters,

and partly (the rest) was transformed to the potential energy of the spring E = {{{(k*d^2)/2}}},

where {{{mu}}} 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

{{{(mv^2)/2}}} = {{{mu*mg*d}}} + {{{(kd^2)/2}}}.

Substitute the given data, and you will get the equation to solve

{{{(1.6*v^2)/2}}} = {{{0.24*186*9.81*4}}} + {{{(1.3*4^2)/2}}}.

Thus the setup is done.

Now solve the equation for "v".

It is just arithmetic.
</pre>