Posted by **help** on Tuesday, November 23, 2010 at 10:02am.

A moving 2.8 kg block collides with a horizontal spring whose spring constant is 215 N/m (see figure). The block compresses the spring a maximum distance of 11.5 cm from its rest position. The coefficient of kinetic friction between the block and the horizontal surface is 0.48

How much mechanical energy is being dissipated by the force of friction while the block is being brought to rest by the spring?

I found the work done by the spring -1.42J

- physics -
**drwls**, Tuesday, November 23, 2010 at 10:11am
I get the friction work to be

(friction force)(distance moved) =

(0.48)(2.8)(9.8)(0.115) = 1.52 J

You don't need the spring constant to get the answer.

- physics -
**help**, Tuesday, November 23, 2010 at 10:21am
the next part is

What is the speed of the block when it hits the spring?

i did

1.42=.5mv^2(final)-.5mv^2(initail)

The final velocity woulf be zero

so

1.42=.5(2.8)v^2

v=1.01

im not sure what i did wrong

- physics -
**drwls**, Tuesday, November 23, 2010 at 10:31am
Initial KE = (stored spring PE when stopped) + friction work

(1/2)MV^2 = (1/2)kX^2 + 1.52 J

Now recalculate V. Include the units. Numbers by themselves, without units, are useless.

- physics -
**Anonymous**, Tuesday, November 23, 2010 at 12:07pm
thanks

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