4
$\begingroup$

First of all, I wonder: in $F=ma$ does the acceleration have to be constant? I believe so but, just as confirmation.

Problem:

A 4.80 kg bucket of water is accelerated upward by a cord of negligible mass whose breaking strength is 75 N. If the bucket starts from rest, what is the minimum time required to raise the bucket a vertical distance of 12 m without breaking the cord?

Thought 1.
I just put in 75N, and and the mass into $f=ma$, and I get the maximum acceleration, which was wrong, so I realized that the bucket itself must exert a force onto the rope.

Putting the mass of the bucket into $ma$, and the gravitational acceleration I got 47.088 newton. 75-47.088=27.912. The rope can only have a force dragging it which is the maximum 27.912 newton before it breaks. Putting this into the formula again $f=ma$, $a=f/m$ to find the maximum acceleration, before the rope breaks. Then I put this into the distance formula. $$ x=x_0+v_0t+\frac12at^2, $$ to solve for the time which was 2.03 seconds.

Was this right? I have to ask for confirmation on this as studying alone in the summer is hard. Often there are only small questions we need answers to, to learn something big.

| cite | improve this question | | | | |
$\endgroup$
  • $\begingroup$ F = ma does not require a to be constant. Also, I'd undo the change to your formatting; the problem is much harder to read now, as a single line with a scroll bar. $\endgroup$ – Harry Wilson Jul 23 '15 at 19:40
  • $\begingroup$ It's good that you tried to solve it yourself and showed your approach. $\endgroup$ – Mike Dunlavey Jul 23 '15 at 20:19
2
$\begingroup$

No, Newton's Second Law of Motion does not require the acceleration to be constant. However, for any given acceleration (and mass) there can only be one value that the force can yield, namely $m \cdot a$.

In other words, if the force and mass stays the same, then the acceleration stays the same as well. Contrary, if the acceleration and mass stays the same, the force stays the same.

Assuming constant acceleration the distance travelled during a certain time period $t$ can be expressed as $v_0 t + \frac{at^2}{2}$, where $v_0$ is the initial velocity. As the initial velocity for your bucket is zero, this simply vanishes to $\frac{at^2}{2}$.

Now, let's think of the maximum acceleration the cord can handle. The force the cord will feel is $(\text{gravitational acceleration} + \text{acceleration}) \cdot \text{mass of bucket}$. We also knew that it would break when the force acceded 75 N, so we look for a value of acceleration where $F = \text{75 N}$. Given that $g = 9.8 \text{m/s}^2$:

$F = \text{75 N} = (g + a) \cdot \text{4.80 kg}$

Solving for $a$ yields $a = \text{5.825 m/s}^2$.

Now that we know the maximum acceleration the cord can handle we put it in $\text{12 m} = \frac{at^2}{2}$ and solve for $t$. Doing that, we get $t = \pm \text{2.03 seconds}$.

So the answer would be $\small{\approx} \text{2 seconds}$.

| cite | improve this answer | | | | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.