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A brick of mass m = 0.21 kg is set against a spring with a spring constant of k1 = 690 N/m which has been compressed by a distance of 0.1 m. Some distance in front of it, along a frictionless surface, is another spring with a spring constant of k2 = 489 N/m.
(A) How far d2, in meters, will the second spring compress when the block runs into it?
(B) How fast v, in meters per second, will the block be moving when it strikes the second spring?
(C) Now assume friction is present on the surface in between the ends of the springs at their equilibrium lengths, and the coefficient of kinetic friction is ?k = 0.5. If the distance between the springs is x = 1 m, how far d2, in meters, will the second spring now compress?

Respuesta :

onyebuchinnaji

The compression of the second spring when the block runs into it is 0.118 m.

The speed of the first block when it collides with the second block is 5.73 m/s.

The compression of the second spring in the presence of friction between the surface of the two springs is 0.099 m.

The given parameters;

  • mass of the brick, m = 0.21 kg
  • first spring constant, k₁ = 690 N/m
  • extension of the first spring, x₁ = 0.1 m
  • second spring constant, k₂ = 489 N/m

(a)

The energy possessed by the first spring is calculated as;

[tex]E_1 = \frac{1}{2} k_1x_1^2\\\\E_1 = \frac{1}{2} \times 690 \times 0.1^2\\\\E_1 = 3.45 \ J[/tex]

Based on the principle of conservation of energy, the second spring will acquire equal energy as the first when it runs into it.

[tex]E_2 = E_1\\\\E_2 = \frac{1}{2} k_2x_2^2\\\\3.45 = \frac{1}{2} \times 489 \times x_2^2\\\\x_2^2 = 0.014\\\\x_2 = \sqrt{0.014} \\\\x_2 = 0.118 \ m\\\\[/tex]

(b)

The speed of the first block when it collides with the second block is calculated as;

[tex]\frac{1}{2} mv^2 = \frac{1}{2} kx^2\\\\mv^2 = kx^2\\\\v= \sqrt{\frac{kx^2}{m} } \\\\v = \sqrt{\frac{690 \times 0.1^2}{0.21} } \\\\v = 5.73 \ m/s[/tex]

(c)

The frictional force present on the surface between the ends of spring will reduce the energy transferred to the second spring.

The compression of the second spring in the presence of friction between the surface of the two springs is calculated as;

[tex]E_1 - F_kx = E_2\\\\\frac{1}{2} k_1x_1^2 - \mu_k mgx= \frac{1}{2} k_2x_2^2\\\\\frac{1}{2} \times 690 \times (0.1)^2 \ - \ 0.5\times 0.21 \times 9.8\times 1 = \frac{1}{2} \times 489\times x_2^2\\\\2.421 = 244.5x_2^2\\\\x_2^2 = \frac{2.421}{244.5} \\\\x_2^2 = 0.0099\\\\x_2 = \sqrt{0.0099} \\\\x_2 = 0.099 \ m[/tex]

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