A rectangular mass M = 10.0 kg is attached to a fixed spring of spring constant

Question

A rectangular mass M = 10.0 kg is attached to a fixed spring of spring constant                  k = 12.0 N/m. The surface on which this mass rests is horizontal and frictionless. A mass m = 1.00 kg is placed on top of the larger mass. The larger mass is then pulled out so that the spring is extended x0 = 3.50 meters beyond the relaxed length of the spring. The mass is released, and during the resulting motion it is observed that the two masses move together as a single unit.

What can you say about the coefficient of static friction between the two masses?

What is this asking? Solve for us? I found it in terms of a but you cant find a? What I tried was Fstatic friction must be greater than Fspring because the boxes dont move. I solvd for Fspring (probably incorrectly)... Where do I go from here? Please dont just throw out a formula for friction coeff and use the information actually given. This is the second time I am posting this question and I dont want to waste questions for this month

Explanation / Answer

M = 10.0 Kg
k = 12.0 N/m
m = 1.0 Kg
xo = 3.50 m

For spring, F = k*x
F = 12.0 * 3.5
F = 42 N

As the mass moves together,
We know, F = (m+M)*a
42 = (10.0+1.0)*a
a = 3.82 m/s^2

Let the coefficient of static friction between the two masses be uk.
Now,
uk*m*g > m*a
uk > a/g
uk > 3.82/9.8
uk > 0.390

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