A bucket of sand is suspended from a tree limb by a spring whoseforce constant (
ID: 1679023 • Letter: A
Question
A bucket of sand is suspended from a tree limb by a spring whoseforce constant (spring constant) is 75 N/m. The bucket and sandinitially had a combined mass of 2.0 kg. When a hole is punchedthrough the bottom of the bucket, the sand begins to run out at arate of 10 grams per second. The bucket, steadily becoming lighter,begins to rise. What is the approximate speed of its ascent?Hint - there could be some calculus involved... or perhapsthere's a non-calculus way to think about it
If someone could work out the problem both ways (calc and non-calcif possible) it would be greatly appreciated. I just need to seehow these types of problems are worked out. It will help me studyfor both my physics final and calc final.
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Explanation / Answer
way 1 : calc we have : kx = mg => k*dx/dt = g*dm/dt => the velocity : V = dx/dt = g/k * dm/dt = 9.8/75 *10x10^-3 = 1.307x10^-3 m/s way 2 : non-calc Assume the velocity is constant => the total force is zero at the time t ,the mass of bucket is m and we have : mg = kx at the time t+dt ,the mass of bucket is m - .dt (is the rate of change of mass = 10x10^-3 kg/s) and we have : (m-dt)g = k(x - V*dt) => -dt*g = -kVdt => V = g/k = 1.307x10^-3m/s
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