Question

SPRING PE PRACTICE • A person uses a 10 kg block to compress a spring (with spring constant 8000 N/m) by 0.1 meters. The block is then released and slides along a frictional ramp (lk = 0.2). How far along the ramp does the block slide before coming to a stop? Ax = ? Hy = 0.2 < 0.2 10 kg ww10t04 370

Answer 1

UK = 0.2 10 kg mgsin37 maman mgcos 37

Consider the figure

$sin37=\frac{H}{d}$

$H=dsin37$

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As the block comes to a rest, the elastic potential energy of the spring is converted into potential energy of the block.

Some portion elastic potential energy of the spring is lost due to friction.

$PE_{spring}-W_{friction}=PE_{block}$

$PE_{spring}-F_{friction}*d=PE_{block}$

$PE_{spring}-\mu_{k}N*d=mgH$

$\frac{1}{2}kx^{2}-\mu_{k}*mgcos37*d=mgdsin37$

$0.5*8000N/m*(0.1m)^{2}-0.2*10kg*9.81m/s^{2}*cos37*d=10kg*9.81m/s^{2}*d*sin37$

$0.5*8000*0.1^{2}-0.2*10*9.81*cos37*d=10*9.81*d*sin37$

$40-15.67*d=59.04*d$

$40=59.04*d&plus;15.67*d$

$40=74.71*d$

$\frac{40}{74.71}=d$

ANSWER: ${\color{Red} d=0.5354m}$

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