www.webassign.net/web/ Set and solve r w LEARN MORE REMARKS Friction or drag fro

Question

www.webassign.net/web/ Set and solve r w LEARN MORE REMARKS Friction or drag from Immersion in a fluld damps the mation of an spring, eventunlly brinalng the object to rest I objlect attached to a QUESTION In the case of friction, what percent of the mass first reached the equllibrium point7 (Himt use the ancwers to parts (a) and (c) mechanlcal eneray was lost by the time the Use the worked example above to heip you solve this problem. A block with mass of k attached to hortzontal spring with spring constant k-3.92 × 10: N/m, as shown in the naure. The surtce the block rests upon ts metlonless. e biockts pused out to,-0.0470 m and released. (a) Find the speed of the block at the equllfbrium point. m/s (b) Find the speed when-0.02 m. (c) Repest part (a) it triction acts an the block, with coemcient a-3 EXERCISE HINTS GETING STARTED MTuck Use the values from PRACTICE IT to heip you work this exercise: Suppose the spring system in the last example starts at iclex to the rart, sat has an initiai speed of O82 VS 0and the attached object is given (e) What distance from the origin does the oplect traver before coming to reat. sisuming the surface is frictioniess () How coes the answer change itche cpemcient of köhetie micio 9 4307 (Use the quadratic tormue.)

Explanation / Answer

a)

m = mass of block = 5.53 kg

k = spring constant = 392 N/m

xf = stretch in the spring = 0.0470 m

xo = stretch in spring at equilibrium = 0 m

Vf = initial speed at the time of release = 0 m/s

Vo = speed at equilibrium = ?

Using conservation of energy

(0.5) m Vf2 + (0.5) k xf2 = (0.5) m Vo2 + (0.5) k xo2

(5.53) (0)2 + (392) (0.047)2 = (5.53) Vo2 + (392) (0)2

Vo = 0.396 m/s

b)

x = 0.02 m

V= speed at x = 0.02 m ,

Using conservation of energy

(0.5) m Vf2 + (0.5) k xf2 = (0.5) m V2 + (0.5) k x2

(5.53) (0)2 + (392) (0.047)2 = (5.53) V2 + (392) (0.02)2

V = 0.358 m/s

c)

m = mass of block = 5.53 kg

k = spring constant = 392 N/m

xf = stretch in the spring = 0.0470 m

xo = stretch in spring at equilibrium = 0 m

Vf = initial speed at the time of release = 0 m/s

Vo = speed at equilibrium = ?

uk = coefficient of friction = 0.130

normal force on the block is given as

Fn = mg

kinetic ffrictional force is given as

fk = uk Fn = uk mg

W = work done by frictional force = fk (Xf - Xo) Cos180 = - uk mg  (Xf - Xo)

Using conservation of energy

(0.5) m Vf2 + (0.5) k xf2 - uk mg  (Xf - Xo) = (0.5) m Vo2 + (0.5) k xo2

(5.53) (0)2 + (392) (0.047)2 - (0.130) (5.53 x 9.8) (0.047 - 0) = (5.53) Vo2 + (392) (0)2

Vo = 0.311 m/s

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.