what is the total time to leave the lake?

Question

While sleeping, you were transported to the surface of a frozen lake. When you wake up, you are 34.3m from the nearest shore. The ice is so slippery (without friction) that you cannot walk. You realize that you can use Newton's third law to your advantage and decide to toss the heaviest object you have, a boot, in order to get yourself moving. Take your weight as 570N. you throw your 1.04kg boot with an average force of 254N and the launch takes 0.762s (this is the time interval during which you apply the force). How long does it take for you to reach the margin, including the launch time?

I applied the following, (Average force) $\times$ (time) = $(∆m)\times(∆v)$, so I found the speed from that. After that, I applied, $S = S_0 + vt$, and I found the time, and added it to the launch time, so I found the total time = $0.946s$. but I am not convinced, besides not having used the weight of the person given in the statement

Answer 1

1. From Newton's Third Law, the boot pushes you with a constant force of same magnitude, $254N$, for the same interval of time, that is, $0.762s$. You know your weight, hence you know your mass, so from Newton's Second law you can calculate the constant acceleration that you will undergo for the time interval of $0.762s$.

2. You can also find out the velocity attained at the end of this interval, which would remain constant afterwards, and also the distance covered under the constant acceleration, from simple kinematics equations.

3. Then subtract this distance from $34.3m$, and divide by the velocity you obtained, to get time traveled under constant velocity. Add $0.762s$ to this to get your final answer.

Can you finish now?

Answer 2

The problem is easiest viewed as one of impulse. Impulse is the change in momentum caused by a force acting over an interval of time. It is often used in circumstances where the time interval is very short, and the force (often called the "average force" as it is neither feasibly nor strictly necessary to measure the fluctuation in applied force over such a small time interval) is relatively high.

The impulse $J = F\Delta t = (254)(0.762) = 193.548 \mathrm{Ns}$, which is the momentum of the boot immediately upon release.

Your initial momentum was zero, so by conservation of linear momentum (no external force), the sum of the momenta of the boot and the "rest of you" will also be zero. Which means the rest of you will be imparted a momentum of $193.548 \mathrm{Ns}$ directly opposite the direction of the momentum of the launched boot, relative to the rest frame of the frozen lake.

Your initial weight $W = mg$, which means your initial mass can be computed as $\frac{570}{9.81} \approx 58.104 \mathrm{kg}$. Subtract the mass of the boot, and the "rest of you" has the mass $57.064 \mathrm{kg}$.

So the rest of you has the momentum of $193.548 \mathrm{Ns}$ imparted to it, which means your initial velocity is $\frac{193.548}{57.064} \approx 3.392\mathrm{m/s}$ directly away from the launch direction of the boot (and directly toward the near edge of the lake). Since there is no resistance to motion (no friction, no air resistance) or other forces directly affecting the momentum in this plane, there is no further acceleration and you can easily compute the time using $s = vt$

The time taken to reach the margin is $\frac{34.3}{3 392} \approx 10.113 \mathrm{s}$ from the time of launch. Adding the launch time and using the appropriate $3$ significant figure precision, the final answer is $10.9 \mathrm{s}$.