Question

A body of mass m = 18.6 kg is attached to a spring with force constant k = 14.3 N/m. This body, which is initially at its equilibrium position, is given an initial velocity of v = 7.8 m/s. What is the speed of this body when it is at position x = - 2.0 m?

An object starting at rest rotates with constant angular acceleration α = 0.3 rad/s² . What is the angular displacement after t = 3.0 s?

Answer 1

Solution :

Question (1) Solution :

Given :

m = 18.6 kg

k = 14.3 N/m

v = 7.8 m/s

x = - 2 m

.

According to conservation of energy : PE_spring + KE = KE_initial

∴ (1/2) k x² + (1/2) m (v_f)² = (1/2) m v²

∴ k x² + m (v_f)² = m v²

∴ (14.3 N/m)(- 2 m)² + (18.6 kg) (v_f)² = (18.6 kg)(7.8 m/s)²

∴ (v_f)² = 57.765 m²/s²

∴ v_f = 7.6 m/s

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Question (2) Solution :

Given :

α = 0.3 rad/s²

Initial angular velocity (ω₀) = 0 rad/s

t = 3 s

.

Let the angular displacement of the object be θ.

Then, Using equation : θ = ω₀ t + (1/2) α t²

∴ θ = (0 rad/s)(3 s) + (0.5)(0.3 rad/s²)(3 s)²

∴ θ = ​​​​​​​1.35 rad = 77.35 degrees