Force given mass, velocity, and displacement

In summary, a 6.0 g bullet leaves the muzzle of a rifle with a speed of 338 m/s. The force exerted on the bullet while it is traveling down the 0.9 m long barrel of the rifle is .006kg (380.8N).
  • #1
mandy9008
127
1

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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  • #2
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.
 
  • #3
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.
 
  • #4
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?
 
  • #5
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?
 
  • #6
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.)
 

What is force?

Force is a physical quantity that describes the amount of push or pull applied to an object.

How is force related to mass, velocity, and displacement?

According to Newton's Second Law of Motion, force is equal to the product of an object's mass and acceleration. Velocity and displacement are both components of acceleration, so they also play a role in determining the force acting on an object.

What units are used to measure force?

In the International System of Units (SI), force is measured in Newtons (N). In the imperial system, force is measured in pounds (lb).

How does increasing mass affect force?

As mass increases, force also increases. This means that a greater force is needed to accelerate a heavier object compared to a lighter object.

What is the relationship between force and displacement?

The relationship between force and displacement is described by the concept of work. Work is equal to the product of force and displacement, and is a measure of the energy transferred to or from an object. Therefore, the greater the displacement, the more work is done, and the greater the force applied.

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