The following system is created to measure the coefficient of kinetic friction o

Question

The following system is created to measure the coefficient of kinetic friction of a given surface material. A 15 kg block is pressed against a spring with spring coefficient of 1400N/m and an initial compression of 35.0 cm. (Figure 1)

In this example, the block will rest against spring #1 initially and is then released. Spring #1 will decompress and shoot the block across the frictional section to spring #2 which temporarily compresses and then decompresses. The block then returns across the frictional section a second time and finally compresses spring #1 again.

If the block compresses spring #1 only 30.5cm on the return trip, find the coefficient of kinetic friction for the 1.00 meter section of material. Spring #2 is just there to change the direction of the block at the other end and does no net work.

Explanation / Answer

energy lost = .5k(x1^2 - x2^2 ) = 0.5 1400*(0.1225 - 0.930) = 20.6325 joules

the energy lost will be due to friction force

hence

F*s = 20.6325

uM*g *s = 20.6325

u = 20.6325/150 = 0.13755

hence coefficient of kinetic friction = 0.13755

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