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A block of mass =4.20 kg slides along a horizontal table with velocity 0=3.50 m/s . At =0 , it hits a spring with spring constant =26.00 N/m and it also begins to experience a friction force. The coefficient of friction is given by =0.250 . How far has the spring compressed by the time the block first momentarily comes to rest? Assume the positive direction is to the right.

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onyebuchinnaji

The distance in which the spring is compressed is 1.065 m.

The given parameters;

  • mass of the block, m = 4.2 kg
  • velocity of the block, v = 3.5 m/s
  • spring constant, k = 26 N/m
  • coefficient of friction, μ = 0.25

Apply the principle of conservation of energy to determine the distance in which the spring is compressed.

[tex]K.E = W + U\\\\\frac{1}{2} mv^2 = \mu Fx \ + \ \frac{1}{2} kx^2\\\\\frac{1}{2}(4.2)(3.5)^2 = 0.25(4.2 \times 9.8)(x) \ + \frac{1}{2} (26)(x^2)\\\\25.725 = 10.29x \ + \ 13x^2\\\\13x^2 + 10.29x - 25.725 = 0\\\\[/tex]

solve the quadratic equation using formula method;

a = 13, b = 10.29,  c = -25.725

[tex]x = \frac{-b \ \ +/- \ \sqrt{b^2 - 4ac} }{2a} \\\\x = \frac{-(10.29) \ \ +/- \ \sqrt{(10.29)^2 \ - \ 4(13\times 25.725)} }{2(13)} \\\\x = 1.065 \ m, \ \ or \ -1.86 \ m[/tex]

choose positive distance, x = 1.065 m.

Thus, the distance in which the spring is compressed is 1.065 m.

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