Physics, 26.07.2019 12:00, rebeccamckellpidge

Astudent uses a compressed spring of force constant 22 n/m to shoot a 0.0075 kg eraser across a desk. the magnitude of the force of friction on the eraser is 0.042 n. how far along the horizontal desk will the eraser slide if the spring is initially compressed 0.035m

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The initial elastic potential energy stored in the spring is:
$U= \frac{1}{2}k x^2 = \frac{1}{2}(22 N/m)(0.035 m)^2=0.0135 J$

Then, the spring is released, all this potential energy is converted into kinetic energy of the eraser:
U= K=0.0135 J

Then, the force of friction does a work to stop the eraser in this motion, and this work is equal to
W=Fd

where F is the magnitude of the friction force and d is the distance covered by the eraser.

For energy conservation, the work done by the friction force must be equal to the initial energy of the eraser, so:
K=W=Fd

and so we find d:
$d= \frac{K}{F}= \frac{0.0135 J}{0.042 N}=0.32 m$

answered: Guest

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answered: Guest

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