Badmouton
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1.A block of mass M=1 kg rests on a frictionless surface and is connected to a horizontal spring of force constant k=25 N/m. The other end of the spring is attached to a wall. A second block of mass m=500 g, rests on top of the first block. The coefficient of static friction between the blocks is 0.3. Find the maximum amplitude of oscillation such that the top block will not slip on the bottom block.
2.These are the equations I used:
F=ma
a(t)=-w2Acos(wt+FI)
w=sqrt(k/m)
Fs=u*normal force
3. Since we don't want the upper block to move, then the force acting on it, namely the spring bobbing under it, must be lower or equal to Fs. This means that the mass*a=Fs. Since I already have the value of the mass (M+m=1kg+0.5g=1.5kg), all I have to find is the acceleration. I calculated the value of w to be 4.08 and the value of Fs to be 1.4. With the acceleration found the following way F=Fs, I derived 1.5kg*a=1.4, thus a=0.93, I can us this equation to isolate A "a(t)=-w2Acos(wt+FI)", since I want the max acceleration, I will assume cos(wt+FI)=1.
Finally this leads me to this equation 0.93=-(4.08)2A, thus A=0.056m... However, the right answer is 0.1764m, can someone help?
2.These are the equations I used:
F=ma
a(t)=-w2Acos(wt+FI)
w=sqrt(k/m)
Fs=u*normal force
3. Since we don't want the upper block to move, then the force acting on it, namely the spring bobbing under it, must be lower or equal to Fs. This means that the mass*a=Fs. Since I already have the value of the mass (M+m=1kg+0.5g=1.5kg), all I have to find is the acceleration. I calculated the value of w to be 4.08 and the value of Fs to be 1.4. With the acceleration found the following way F=Fs, I derived 1.5kg*a=1.4, thus a=0.93, I can us this equation to isolate A "a(t)=-w2Acos(wt+FI)", since I want the max acceleration, I will assume cos(wt+FI)=1.
Finally this leads me to this equation 0.93=-(4.08)2A, thus A=0.056m... However, the right answer is 0.1764m, can someone help?