We know that, Etotal = K.Eblock + P.Espring
Etotal = (1/2) m v02 + (1/2) k x02
Etotal = [(0.5) (0.14 kg) (0.18 m/s)2] + [(0.5) (3 N/m) (-0.049 m)2]
Etotal = [(0.002268 J) + (0.0036015 J)]
Etotal = 0.0058695 J
At the maximum amplitude, the velocity of a block will be zero.
Then, we have
E = (1/2) k A2
(0.0058695 J) = (0.5) (3 N/m) A2
(0.0058695 J) = (1.5 N/m) A2
A2 = [(0.0058695 J) / (1.5 N/m)]
A = 0.003913 m2
A = 0.0625 m
Part B : What is the block's maximum acceleration?
From an obey's Hooke law, we get
Fmax = k A m amax
amax = k A / m
amax = [(3 N/m) (0.0625 m)] / (0.14 kg)
amax = [(0.1875 N) / (0.14 kg)]
amax = 1.34 m/s2
Part C : What is the block's position when the acceleration is maximum?
We know that, x = Amax
Then, we get
x = 0.0625 m
Converting m into cm :
x = 6.25 cm
Part D : What is the speed of a block when x1 = 3.5 cm?
Using conservation of energy, we have
(K.Eblock + P.Espring)initial = (K.Eblock + P.Espring)final
(1/2) m v02 + (1/2) k x02 = (1/2) m vf2 + (1/2) k xf2
m v02 + k x02 = m vf2 + k xf2
[(0.14 kg) (0.18 m/s)2 + (3 N/m) (-0.049 m)2] = [(0.14 kg) vf2 + (3 N/m) (0.035 m)2]
(0.011739 J) = (0.14 kg) vf2 + (0.003675 J)
[(0.011739 J) - (0.003675 J)] = (0.14 kg) vf2
(0.008064 J) = (0.14 kg) vf2
vf2 = [(0.008064 J) / (0.14 kg)]
vf = 0.0576 m2/s2
vf = 0.24 m/s
Converting m/s into cm/s :
vf = 24 cm/s
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