Answer:
(a) The block's initial mechanical energy is 10.0125 J
(b) The distance the block travel before coming to rest is 0.29 meter
Explanation:
m= 1.5kg, k - 890N/m, θ = 22°, d = 0.15m, μk = 0.15
Initial velocity, u = 0
Initial gravitational energy, [tex]U_{go}[/tex] = 0
a) Initial Mechanical Energy
E₀ = U₅₀ + [tex]U_{go}[/tex] + K₀
Block at rest, so K₀ = Kinetic energy = 0
E₀ = 1/2 kd²
= 0.5 x 890 x 0.15²
E₀ = 10.0125 J
The block comes to rest again after moving a distance 'L'
[tex]U_{fs} = \frac{1}{2} k(L-d)^{2}[/tex]
[tex]U_{gf} = mgh = mgL.Sin[/tex] θ
From energy conservation
[tex]E_{i} = E_{f}[/tex]
10.0125 = 1/2(890) (L-0.15)² + (1.5)(98)L.Sin22⁰
10.0125 = 445L² - 133.5L + 10.0125 + 5.51L
445L² - 127.99L = 0
==> L = 0 or L = 0.29 meter