A 0.535 kg wood block is firmlyattached to a very light horizontal spring ( k =1

Question

A 0.535 kg wood block is firmlyattached to a very light horizontal spring (k =165 N/m) as shown in Fig. 6-40. It isnoted that the block-spring system, when compressed 5.0 cm andreleased, stretches out 2.3 cm beyond the equilibrium positionbefore stopping and turning back. What is the coefficient ofkinetic friction between the block and the table? A 0.535 kg wood block is firmlyattached to a very light horizontal spring (k =165 N/m) as shown in Fig. 6-40. It isnoted that the block-spring system, when compressed 5.0 cm andreleased, stretches out 2.3 cm beyond the equilibrium positionbefore stopping and turning back. What is the coefficient ofkinetic friction between the block and the table?

Explanation / Answer

The difference between the initial energy (before the block isreleased) and the final energy (at the point where it stops) tells us the energy lost due to friction. This energy is equal to the work done by the force W=Fd where d is the distance traveled and F is the frictional forcegiven by F=N where N is the normal force and the coefficient of kineticfriction. The energy of a spring is E= 1/2 k x2 where x is the distance from the equilibrium point xi = 0.05 m Ei = 0.20625 J xf= 0.023 m Ef = 0.0436425 so the energy lost due to friction is W = Ei - Ef = 0.1626075 F = mg m=0.535 Kg g = 9.8 ms-2 d = 5cm + 2.3cm = 0.073 m so = W/(mgd) = 0.425

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