(a) Find the speed of the block at the equilibrium point. m/s (b) Find the speed

Question

(a) Find the speed of the block at the equilibrium point.
  m/s

(b) Find the speed when x = 0.02 m.
  m/s

(c) Repeat part (a) if friction acts on the block, with coefficient k = 0.170.
  m/s

EXERCISE

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 of0.56 m/s.

(a) What distance from the origin does the object travel before coming to rest, assuming the surface is frictionless?
m

(b) How does the answer change if the coefficient of kinetic friction is k = 0.170? (Use the quadratic formula.)
m

Explanation / Answer

(1) m = 5.02 kg, k =4.43x10^2 N/m , xi =0.046 m

(a) xi =0.046

From conservaiton of energy

Ki+Ui =Kf +Uf

(1/2)mv2+0 = 0 + (1/2)kx2

v = x [k/m]1/2 = 0.046[443/5.02]1/2

v = 0.432 m/s

(b) xi =0.046 m, x =0.02 m

From conservaiton of energy

Ki+Ui =Kf +Uf

(1/2)mv2+(1/2)kx2 = 0 + (1/2)kxi2

v2 = (k/m) [xi2 - x2]

v2 = (443/5.02)[(0.046)2 -(0.02)2]

v = 0.389 m/s

(c) k =0.17, xi =0.046 m

From work energy theorem

W = Kf -Ki

-Fk .xi = -(1/2)mv2

k mg xi =(1/2)mv2

v2 = 2k g xi = 2*0.17*9.8*0.046

v = 0.39 m/s

Exercise:

(a) v =0.56 m/s

From conservaiton of energy

Ki+Ui =Kf +Uf

(1/2)mv2+0 = 0 + (1/2)kx2

x = v [m/k]1/2 = 0.56[5.02/443]1/2

x = 0.0596 m

(b)

k =0.17, v =0.56 m/s

From work energy theorem

W = Kf -Ki

Fk .xi = -(1/2)mv2 =(1/2) kx2

k mg xi =(1/2)kxi2

xi2 = 2k g m/kxi = (2*0.17*9.8)/(443*0.0596)

xi = 0.36 m/s

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