Newton's Second Law — Net Force problem

In summary, the work done on the potato is related to its kinetic energy. The calculations appear to be correct. Good job.
  • #1
adams_695
16
1
Homework Statement
A group of budding engineers made a cannon 1.0 m in length that shoots potatoes. They recorded the speed at which a potato could be launched. They measured it to be a constant 63 m/s over a horizontal distance of 1.0 m. The potato had a mass of 174 g. Assuming the potato accelerated from rest in the cannon, what was the magnitude of the force that the cannon exerted of the potato? (assume the potato hits a wall after 1.0 m and stops moving) (Answer to one decimal place)
Relevant Equations
Newton's Second Law/Third Law
working of Q16 2.png
working of Q16 2.png
 
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  • #2
Posting your work sideways is bad form
 
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  • #3
phinds said:
Posting your work sideways is bad form

edited.
 
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  • #4
phinds said:
Posting your work sideways is bad form

is the working correct from what you can see?
 
  • #5
adams_695 said:
is the working correct from what you can see?
From what you have written you are relating the work done on the potato to its kinetic energy. That is correct if you are assuming that the friction in the cannon barrel is negligible. Since that is not stated you should mention it (you can get bonus marks for noticing things like that!). Your calculations appear to be correct. Good job.

AM
 
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  • #6
adams_695 said:
is the working correct from what you can see?
The question is flawed. It ought to tell you to assume the force applied is constant. If it is not then you do not have enough information to find the average force.
 
  • #7
I'm just a biologist, but it doesn't look right to me. Close but no cigar. This is a time:distance problem and the correct SI units is kg not g. Thus:
average velocity: (63 m/s - 0 m/s)/2 = 31.5 m/s
time to terminal velocity: 1 m/31.5 m/s=.032s
acceleration to go 1 meter: 1 m= a/2 m/s^2 * t^2: rearranged a= 2*1/.032^2=1953.1 m/s^2
F=ma: F=0.174 kg * 1953.1 m/s^2 = 339.8 kgm^2 = 339.8 N
 
  • #8
gingerjake said:
I'm just a biologist, but it doesn't look right to me. Close but no cigar. This is a time:distance problem and the correct SI units is kg not g. Thus:
average velocity: (63 m/s - 0 m/s)/2 = 31.5 m/s
time to terminal velocity: 1 m/31.5 m/s=.032s
acceleration to go 1 meter: 1 m= a/2 m/s^2 * t^2: rearranged a= 2*1/.032^2=1953.1 m/s^2
F=ma: F=0.174 kg * 1953.1 m/s^2 = 339.8 kgm^2 = 339.8 N
Your method ought to have given exactly the same answer, 345.3N. Since it did not, your more circuitous route must have introduced some rounding or arithmetic error. Try keeping more decimal places.
 
  • #9
haruspex said:
The question is flawed. It ought to tell you to assume the force applied is constant. If it is not then you do not have enough information to find the average force.
As an object accelerates it takes more power to generate a constant acceleration. These questions are doubly flawed, therefore, as the assumption of constant acceleration is unlikely to be valid.
 
  • #10
I'm lucky I even got an answer. U R right, I truncated the 1/31.5 quotient before I squared it. Should have kept all 11 digits. {:-)
 
  • #12
PeroK said:
As an object accelerates it takes more power to generate a constant acceleration. These questions are doubly flawed, therefore, as the assumption of constant acceleration is unlikely to be valid.
It would just require a source of constant pressure gas applied to the chamber. A large compressor could do that quite easily.

AM
 
  • #13
Andrew Mason said:
It would just require a source of constant pressure gas applied to the chamber. A large compressor could do that quite easily.

In practice, no projectile launchers, including potato cannons work this way, because it's not really easy. A large compressor is not enough, one needs to design a system with adequate flow from the compressor to the chamber and also work out the details so that the flow is just right to maintain constant pressure. Most projectile launchers that use a compressed gas allow a fixed reservoir of gas to simply expand as the projectile is pushed in the barrel. The designs that are closest to having a constant pressure (thus a constant force) achieve this simply by having the reservoir volume large enough so that the final volume (reservoir plus barrel volume) is only a smidge larger than the initial volume (reservoir volume). But such a large reservoir volume introduces other issues of cost and size.

The other main approach is using combustion designs. Most of these reach a peak pressure before the projectile is half way down the tube and then the pressure drops.
 
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  • #14
haruspex said:
The question is flawed. It ought to tell you to assume the force applied is constant. If it is not then you do not have enough information to find the average force.
It is an odd question. It omits necessary assumptions and it adds an unnecessary assumption about what happens to the potato after launch.

AM
 
  • #15
adams_695 said:
is the working correct from what you can see?
I think your solution is equivalent to one of a list of perhaps 2 or 3 methods that would get full credit.

One quibble about your worksheet: in one line you gave g as the unit where I think you knew, from how you converted the numerical part, is kg. That could cost you.

There were several comments about finer details, that were true, but I expect they were intended to be neglected. For example, in projectile motion problems, they often do not tell you to ignore air drag, but then give you no info regarding how to consider air drag.
 

FAQ: Newton's Second Law — Net Force problem

1. What is Newton's Second Law?

Newton's Second Law states that the acceleration of an object is directly proportional to the net force acting on the object and inversely proportional to its mass. This can be expressed as F=ma, where F is the net force, m is the mass of the object, and a is the acceleration.

2. How is Newton's Second Law used to solve net force problems?

To solve a net force problem, you must first identify all the forces acting on the object. Then, use Newton's Second Law to calculate the net force by adding up all the individual forces. Finally, use the net force and the mass of the object to calculate the acceleration.

3. What is the unit of force in Newton's Second Law?

The unit of force in Newton's Second Law is Newton (N). One Newton is equal to 1 kilogram meter per second squared (kg m/s²).

4. Can Newton's Second Law be applied to objects with varying mass?

Yes, Newton's Second Law can be applied to objects with varying mass. This is because the acceleration is inversely proportional to the mass, meaning that as the mass increases, the acceleration decreases, and vice versa.

5. What is the difference between mass and weight in Newton's Second Law?

Mass and weight are often used interchangeably, but they are actually two different concepts in Newton's Second Law. Mass is a measure of an object's inertia, while weight is a measure of the force of gravity acting on an object. In other words, mass is constant, while weight can vary depending on the strength of the gravitational pull on the object.

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