Force given mass, velocity, and displacement

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A 6.0 g bullet exits a rifle at 338 m/s, and the problem involves calculating the constant force exerted on it while traveling down a 0.9 m barrel. The initial approach using x = vt is incorrect due to the assumption of constant velocity, as the bullet experiences constant acceleration. Instead, applying the kinematic equation vf² = vo² + 2ax allows for the calculation of acceleration, leading to a force of approximately 380.8 N. The discussion emphasizes the importance of using the correct equations for scenarios involving acceleration and suggests considering the work-energy theorem for a more straightforward solution. Understanding the relationship between work, force, and energy is crucial for solving such physics problems effectively.
mandy9008
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Homework Statement


A 6.0 g bullet leaves the muzzle of a rifle with a speed of 338 m/s. What force (assumed constant) is exerted on the bullet while it is traveling down the 0.9 m long barrel of the rifle?


Homework Equations


F=ma
x=vt
a=v/t

The Attempt at a Solution


x=vt
.9m=338m/s t
t=.0027s

a=v/t
a=338 m/s /.0027s
a=126937.8 m/s2

F=ma
F=.006kg (126937.8 m/s2)
F=761.6N
 
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You cannot use x = vt. It is valid only when the acceleration is zero - not the case here. Instead, start with the kinematic equation that relates displacement, acceleration and speed and has no explicit time dependence.
 
First let me just ask you a question. Is your context now up to energy (or example, conservation of energy)?

For this problem you are giving the length of the path, or the height.
And you have the final speed.

Using conservation of energy you will be able to find the the initial speed. This will allow you to compute the force using work-kinetic energy theorem which involves velocity (in this case our initial velocity).

However, using your approach. It can be done too.

What you need didn't take into account is exactly the notion of "initial speed".
Have your initial speed compute, and use whatever you have, find the acceleration, and finally get the force.
 
v^2=Vo^2 + a x is this right?
338^2 m/s=a (.9m)
a=63468.9 m/s^2

F=ma
F= .006kg (63468.9m/s^2)
F=380.8N

correct?
 
mandy9008 said:

The Attempt at a Solution


x=vt
.9m=338m/s t
t=.0027s

Sorry, but the above 'x = vt' equation only applies to objects moving at a constant velocity. This problem assumes a constant force (thus constant acceleration) which is quite different. Your value for t isn't correct.

If you must use kinematics to solve this problem, try one of your other equations that assume a constant acceleration. (It is possible to solve this problem using kinematics but there is an easier way [see below]).

But I should point out that there is an easier way to solve this problem using the work-energy theorem; a way of expressing conservation of energy. Here are some questions,
(a) What is the definition of work, W, done by a constant force over a given distance?
(b) What is the definition of kinetic energy, given an object's mass and velocity?
 
mandy9008 said:
v^2=Vo^2 + a x is this right?

You left out a '2', but you got the right answer below so I assume it was just a typo. Just to be clear, the equation is,

vf2 = v02 + 2ax

338^2 m/s=a (.9m)
a=63468.9 m/s^2

F=ma
F= .006kg (63468.9m/s^2)
F=380.8N

correct?

Looks okay to me :approve:

(ps. if you are up to the point in your coursework where are are studying work and conservation of energy, I suggest you solve it that way too. If not, you have something to look forward to.)
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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