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A spring lies on a horizontal table, and the left end of the spring is attached to a wall. The other end is connected to a box. The box is pulled to the right, stretching the spring. Static friction exists between the box and the table, so when the spring is stretched only by a small amount and the box is released, the box does not move. The mass of the box is 0.50 kg, and the spring has a spring constant of 68N/m. The coefficient of static friction between the box and the table on which it rests is μs 0.52. How far can the spring be stretched from its unstrained position without the box moving when it is released?

Sagot :

Answer:

the maximum distance = 3.75 cm

Explanation:

To find the maximum distance the spring can be stretched without the box moving when it is released, we have to find out all the forces that work on the box.

Refer to the picture, there are 4 forces work on the box:

  • [tex]F_{spring}=\text{spring force}[/tex]
  • [tex]f_s=\text{static force}[/tex]
  • [tex]w=\text{weight}[/tex]
  • [tex]N=\text{normal force}[/tex]

Given:

  • [tex]\text{mass (m)}=0.50\ kg[/tex]
  • [tex]\text{spring constant (k)}=68\ N/m[/tex]
  • [tex]\text{coefficient of static friction }(\mu_s)=0.52[/tex]

By applying these formulas, we can find the distance:

[tex]\boxed{w=mg}[/tex]

[tex]\boxed{f_s=\mu_sN}[/tex]

[tex]\boxed{F_{spring}=kx}[/tex]

Since the box is in an equilibrium state (not moving), then:

  • [tex]\Sigma F_y=0[/tex]
  • [tex]\Sigma F_x=0[/tex]

[tex]\Sigma F_y=0[/tex]

[tex]N-w=0[/tex]

[tex]N=w[/tex]

[tex]N=mg[/tex]

[tex]N=0.50\times9.8[/tex]

[tex]N=4.9\ N[/tex]

[tex]\Sigma F_x=0[/tex]

[tex]F_{spring}-f_s=0[/tex]

[tex]kx-\mu_sN=0[/tex]

[tex]68x-0.52\times4.9=0[/tex]

[tex]x=2.548\div68[/tex]

[tex]x=0.0375\ m[/tex]

[tex]\bf x=3.75\ cm[/tex]

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