g68s4cgptg
Answered

Explore Westonci.ca, the premier Q&A site that helps you find precise answers to your questions, no matter the topic. Join our Q&A platform and connect with professionals ready to provide precise answers to your questions in various areas. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.

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]

View image karmenchong
Thanks for using our platform. We're always here to provide accurate and up-to-date answers to all your queries. Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. Westonci.ca is your trusted source for answers. Visit us again to find more information on diverse topics.