Question

Part B Review What is the block's maximum acceleration? A 140 g block attached to a spring with spring constant 3.0 N/m oscillates horizontally on a frictionless table. Its velocity is 18 cm/s when o --4.9 cm Express your answer to two significant figures and include the appropriate units. cm aas 155 Submit Previous Answers Request Answer Text Incorrect; Try Again; 19 attempts remaining X

Answer 1

We know that, E_total = K.E_block + P.E_spring

E_total = (1/2) m v₀² + (1/2) k x₀²

E_total = [(0.5) (0.14 kg) (0.18 m/s)²] + [(0.5) (3 N/m) (-0.049 m)²]

E_total = [(0.002268 J) + (0.0036015 J)]

E_total = 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 = _$\sqrt{}$ 0.003913 m²

A = 0.0625 m

Part B : What is the block's maximum acceleration?

From an obey's Hooke law, we get

F_max = k A $\Leftrightarrow$ m a_max

a_max = k A / m

a_max = [(3 N/m) (0.0625 m)] / (0.14 kg)

a_max = [(0.1875 N) / (0.14 kg)]

a_max = 1.34 m/s²

Part C : What is the block's position when the acceleration is maximum?

We know that, x = A_max

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.E_block + P.E_spring)_initial = (K.E_block + P.E_spring)_final

(1/2) m v₀² + (1/2) k x₀² = (1/2) m v_f² + (1/2) k x_f²

m v₀² + k x₀² = m v_f² + k x_f²

[(0.14 kg) (0.18 m/s)² + (3 N/m) (-0.049 m)²] = [(0.14 kg) v_f² + (3 N/m) (0.035 m)²]

(0.011739 J) = (0.14 kg) v_f² + (0.003675 J)

[(0.011739 J) - (0.003675 J)] = (0.14 kg) v_f²

(0.008064 J) = (0.14 kg) v_f²

v_f² = [(0.008064 J) / (0.14 kg)]

v_f = _$\sqrt{}$ 0.0576 m²/s²

v_f = 0.24 m/s

Converting m/s into cm/s :

v_f = 24 cm/s