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JulijaS [17]
3 years ago
7

A student uses a compressed spring of force constant 22 N/m to shoot a 0.0075 kg eraser across a desk. The magnitude of the forc

e 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
Physics
1 answer:
muminat3 years ago
3 0
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
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