Friction or drag from immersion in a fluid damps the motion of an object attache

Question

Friction or drag from immersion in a fluid damps the motion of an object attached to a spring, eventually bringing the object to rest. In the case of friction, what percent of the mechanical energy was lost by the time the mass first reached the equilibrium point? Use the worked example above to help you solve this problem. A block with mass of 5.04 kg is attached to a horizontal spring with spring constant k = 3.25 x 10^2 N/m, as shown in the figure. The surface the block rests upon is frictionless. The block is pulled out to x_i = 0.0460 m and released. Find the speed of the block at the equilibrium point. Find the speed when x = 0.033 m Repeat part (a) if friction acts on the block, with coefficient mu_k = 0.120. Use the values from PRACTICE IT to help you work this exercise. Suppose the spring system in the last example starts at x = 0 and the attached object is given a kick to the right, so it has an initial speed of 0.64 m/s. What distance from the origin does the object travel before coming to rest, assuming the surface is frictionless? How does the answer change if the coefficient of kinetic friction is mu_k = 0.120? (Use the quadratic formula.)

Explanation / Answer

from the law of cosnervation of energy

0.5 kx^2 = 0.5 mv^2

mv^2 = 325 * 0.046^2

v^2 = 325* 0.046^2/(5.04)

v = 0.369 m/s

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when x = 0.033m

from the law of cosnervation of energy

0.5 kxf^2 - 0.5 k xi^2 = 0.5 mv^2

mv^2 = 325 * (0.046^2-0.033^2)

v^2 = 325 * (0.046^2-0.033^2)/(5.04)

v = 0.257 m/s

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c. frictional force f = umg

Ff = 0.18 * 5.04* 9.81

Ff = 8.899 Joules

Velocity at equillibrium is zero

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EXERCISE probems :

0.5 mv^2 = 0.5 kx^2

x^2 = 5.04 * 0.64* 0.64/(325)

x = 7.96 cm
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b)0.5mv^2 = 0.5 kx2 + mgx

0.5 * 325 *x^2 + 0.12*5.04 *9.8* x - 0.5*5.04 *0.64^2 = 0

x=0.0635 m

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