Solve Stretched Spring Homework: Find End of Spring on Ruler

  • Thread starter Thread starter 12boone
  • Start date Start date
  • Tags Tags
    Spring
AI Thread Summary
A vertical spring with a spring constant of 58 N/m is positioned next to a ruler at the 15 cm mark. When a 2.0 kg mass is added, the force exerted by the mass is calculated using F = mg, resulting in an additional displacement of the spring. The calculations indicate that the new position of the spring's end can be determined by adding the displacement to the initial 15 cm mark. The correct approach involves using the formula F = kx to find the extension caused by the added mass. Ultimately, the new measurement on the ruler will reflect this extension accurately.
12boone
Messages
21
Reaction score
0

Homework Statement


A spring with 58 hangs vertically next to a ruler. The end of the spring is next to the 15- mark on the ruler. If a 2.0- mass is now attached to the end of the spring, where will the end of the spring line up with the ruler marks?


Homework Equations



I used F=KX should it be mgy=kx? or mgy=1/2kx^2

The Attempt at a Solution



My solution was 32 centimeters and 34 centimeters and those were both wrong. Must use two sig figs.
 
Physics news on Phys.org
12boone said:

Homework Statement


A spring with 58 hangs vertically next to a ruler. The end of the spring is next to the 15- mark on the ruler. If a 2.0- mass is now attached to the end of the spring, where will the end of the spring line up with the ruler marks?

Homework Equations



I used F=KX should it be mgy=kx? or mgy=1/2kx^2

The Attempt at a Solution



My solution was 32 centimeters and 34 centimeters and those were both wrong. Must use two sig figs.

With F=-kx you can know k by the original measurement - assuming that 0 is the 0 force point. This implies that each unit represents 58*g/15 units of force. Adding another 2*g units of force means that you will get an additional 2*g*15/58*g units. This suggest then it will lengthen to 15+(30/58) units.

I'm curious where you got cm out of your description. Or is there a picture you aren't sharing and other units you haven't mentioned?
 
oh the ruler was 15 cm
 
sorry i guess my copy paste didnt work. The 58 is 58 N/m and is the K and it hangs next to a spring that is next to a ruler at 15 cm. If you add 2 kg to it what is the new measured distance?
 
12boone said:
sorry i guess my copy paste didnt work. The 58 is 58 N/m and is the K and it hangs next to a spring that is next to a ruler at 15 cm. If you add 2 kg to it what is the new measured distance?

I still have no idea where the origin is. But 2 kg will move it 2*9.8/58 m.
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top