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anonymous
 4 years ago
EXAM TOMORROW! URGENT HELP!! THANKS :D A spring of stiffness k supports a box of mass M in which is placed a block of mass m. If the system is pulled downward a distance d from the equallibrium position and then released, find the force of eaction between the block and the bottom of the box as a function of time. For what value of d does the block just begin to leave the bottom of the box at the top of the vertical oscillations? NEGLECT air resistance.
anonymous
 4 years ago
EXAM TOMORROW! URGENT HELP!! THANKS :D A spring of stiffness k supports a box of mass M in which is placed a block of mass m. If the system is pulled downward a distance d from the equallibrium position and then released, find the force of eaction between the block and the bottom of the box as a function of time. For what value of d does the block just begin to leave the bottom of the box at the top of the vertical oscillations? NEGLECT air resistance.

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0this is what I have done so far: So for the whole sysem: \[(M+m)x''=kx\] and therefore: \[x''+k/(M+m) x=0\] which means that x(t): \[x(t)= dcos(\sqrt{k/(M+m)} * t)\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Looks good so far, but this equation holds only as long as the block does not leave the bottom of the box. To find the value of d, when this starts to happen, you just need to think about the maximum force in the highest point (which is directly given by hook's law). As soon as this force is bigger than the gravitational force, the block will leave the bottom, since it's accelerated less then the outer box.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0for me i know i have to assume something when the outer box has reached the bottom of its motion, or x=d. But i dont understand the physics of what happens just when the outer box stop, i think when the outer box stops, the smaller block inside experiences the reaction force which is the outcome of the deceleration of the outer box. however, I still do not know how to write this into equations

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Okay, the real question is, from what point on is the force of the spring inverted, because below this point, the force of the spring itself is always upwards and hence, the acceleration of the outer box is always smaller than the free gravitational acceleration, since it's reduced by the force of the spring. So we need to find the equilibrium position of the spring itself, which is above the equilibrium position of the system. This is also fairly simple, since we know that at the equilibrium state of the system the two forces cancel. So setting them both equal (g*(M+m) = k * x), we see that the equilibrium state of the system is x=g*(M+m)/k below the equilibrium point of the spring. So as soon as d is bigger then this, the outer box is accelerated more than the inner box in the highest point and the inner mass will leave the ground. (Acceleration of the outer box is then a = g+k(dx)/M; of the inner box is a = g.)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Now i get when you say what the real question is, and that makes a lot of sense. But are you saying that g(M+m)+kx=mg

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0because if the spring force is always up, and at the turning point it is down which lowers the outer box accelaration, then that difference in force = mg of the small box

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0in the opposite direction so mg

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0and therefore, mg+kx+(M+m)g=0

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.1Suppose for moment that m was glued to M. Then you know that the equation of motion would be \[ y(t) = y_{e} + d \cos\left( \sqrt{\frac{k}{M+m}} t \right) \] where \( y_e \) is the equilibrium position. Now from that equation, you can write down the equation for acceleration: it's the double time derivative: \[ a(t) = \frac{dk}{M+m} \cos\left( \sqrt{\frac{k}{M+m}} t \right) \] At the top of the oscillation therefore, the acceleration is \[ a_{max} = \frac{dk}{M+m} \] Now, if that is greater than gravitational acceleration, \( g \), then you really will need \( m \) to be glued to \( M \), because the mass will be accelerating down faster than \( m \) would have fallen if gravity were acting alone. Therefore ... the answer to the second part of the question is ... you fill in the gap.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[mg=mkd/(M+m)\]Ok, so, the acceleration of the whole is sometimes larger than that of the acceleration of the block. Hence, \[m(gkd/(M+m))=0\] \[mg=mkd/(M+m)\] and \[d=g(M+m)/k\] I have a question though, does this happen when the system of M+m going down only, so that when g is smaller than the acceleration of the whole system, where m falls inside the box.??

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Okay i tried but i don't have NO idea...sorry.
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