A block of mass m = 1.5 kg is attached to a spring with spring constant k = 890 N/m.
The spring is attached to a fixed wall.
The block is initially at rest on an inclined plane that is at an angle of θ = 22 degrees with respect to the horizontal, and the coefficient of kinetic friction between the block and the plane is μk= 0.15.
In the initial position, where the spring is compressed by a distance of d = 0.15 m, the mass is at its lowest position and the spring is compressed the maximum amount. Assume the initial gravitational energy of the block is zero.
(a) What is the block's initial mechanical energy, E0 in Joules?
(b) If the spring pushes the block up the incline, what distance
L, in meters, will the block travel before coming to rest? The spring remains attached to both the block and the fixed wall throughout its motion.

Respuesta :

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