# Two Physics Problems plsss help ?

#### orange_angel

Two Physics Problems .. plsss help ??!!

Q1 :

The figure shows a mass M inside a tube bent in a circle of radius R=1.20 m. M can slide with negligible friction inside the tube. The tube rotates about a vertical axis passing through the center. The mass is in equilibrium at q =56.5 degrees. Calculate the number of revolution the hoop makes in 12.0 s.

THE FIGURE : http://www3.0zz0.com/2006/12/22/10/35335598.gif [Broken]

Q2:

A 8.40 kg mass suspended from a spring with spring constant, k = 800.0 N/m, extends it to a total length of 0.270 m. Find the total length of the spring when a 13.40 kg mass is suspended from it.

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#### Andrew Mason

Homework Helper
Q1 :

The figure shows a mass M inside a tube bent in a circle of radius R=1.20 m. M can slide with negligible friction inside the tube. The tube rotates about a vertical axis passing through the center. The mass is in equilibrium at q =56.5 degrees. Calculate the number of revolution the hoop makes in 12.0 s.

THE FIGURE : http://www3.0zz0.com/2006/12/22/10/35335598.gif [Broken]

Q2:

A 8.40 kg mass suspended from a spring with spring constant, k = 800.0 N/m, extends it to a total length of 0.270 m. Find the total length of the spring when a 13.40 kg mass is suspended from it.
You will have to show what you have done to attempt a solution.

AM

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#### orange_angel

the second question :i don't know how to begin !!!
............
the first question

F=kx, where F is the force applied, k is the spring constant and x is the distance stretched.

k = 800 N/m
x=0.270
F=kx=800*0.270=216N/m

I don't know how to find the total length of the spring ..

[ i never learn physics before .. i have many problems !! I'm trying to be better ..]

#### Doc Al

Mentor
k = 800 N/m
x=0.270
F=kx=800*0.270=216N/m
Careful: 0.270 m is the total length of the spring, not the amount of stretch. Use Hooke's law to figure out how much the spring stretched when the weight was added--this will allow you to calculate the unstretched length.

Hint for question #1: What kind of motion does the mass undergo? What kind of acceleration?

#### orange_angel

ok ..

in question 2 :

maybe i have to do that :

F=mg=13.40*9.81=131.454 N

x=F/k=131.454/800=0.1643m

is it right ??

#### Doc Al

Mentor
maybe i have to do that :

F=mg=13.40*9.81=131.454 N

x=F/k=131.454/800=0.1643m
That's certainly an essential step in solving the problem. What is the meaning of what you have calculated, and how does it relate to the final solution?