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Use the values from PRACTICE IT to help you work this exercise. Suppose the spri

ID: 2191246 • Letter: U

Question

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 0.51 m/s. (a) What distance from the origin does the object travel before coming to rest, assuming the surface is frictionless?
m NEED THIS ONE

(b) How does the answer change if the coefficient of kinetic friction is ?k = 0.160? (Use the quadratic formula.)
m NEED THIS ONE

A block with mass of 4.69 kg is attached to a horizontal spring with spring constant k = 4.35 102 N/m, as shown in the figure. The surface the block rests upon is frictionless. The block is pulled out to xi = 0.0490 m and released. (a) Find the speed of the block at the equilibrium point. m/s CORRECT ANSWER (b) Find the speed when x = 0.03 m. m/s CORRECT ANSWER (c) Repeat part (a) if friction acts on the block, with coefficient ?k = 0.160. m/s CORRECT ANSWER EXERCISE HINTS: 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 0.51 m/s. (a) What distance from the origin does the object travel before coming to rest, assuming the surface is frictionless? m NEED THIS ONE (b) How does the answer change if the coefficient of kinetic friction is ?k = 0.160? (Use the quadratic formula.)

Explanation / Answer

the idea is that the mass is given an initial amount of KE = 1/2 mv^2; if you know the mass and v, you can calculate initial KE as the spring is stretched, some of this KE goes into PE of the stretched spring, in an amount 1/2kx^2 where k is the spring constant and x is the distance displaced when there is friction, some of this energy is dissipated as work done by friction; the work done by friction is W= f x where f is the frictional force and x is the distance through which the friction acts, equal to the distance the object moves since f= u N = u mg we know that when the object stops, all its initial KE has gone into stretching the string and into work done by friction, so we have 1/2 mv^2 = 1/2 kx^2 + u mg x this is the quadratic equation 1/2 kx^2 + u mg x - 1/2 mv^2=0 solve this for x where the coefficients are a=1/2 k, b=umg, c=-1/2mv^2

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