Mechanics question with variable force

In summary, the conversation discusses a particle of mass m moving in a straight line under the action of a force F, where F = ms. When the displacement s is equal to -a, the velocity of the particle is u. The question is to find the velocity when s = 0. The conversation also touches on the equations F = ma and F = -ma, as well as the relationship between s, v, and a. The participants also discuss the use of a constant, C, in the equations and the possibility of the force being either towards or away from O. Ultimately, there is confusion and uncertainty about the correct formula and whether a given constant was meant to be a or k.
  • #1
Sink41
21
0
Question copied word for word

1. A particle of mass m moves in a straight line under the action of force F where F = ms, s being the displacement of the particle from O, a fixed point on the line.
When s = -a the velocity of the particle is u. Find the velocity of the particle when s = 0.


2. F = ma or F= -ma (i think...)

a = v(dv/ds)


3. If ms = -ma, then s is always equal to -a, so the velocity would always be u, which doesn't look right... Or if I intergate:

s = -a

s = -v(dv/ds)

s^2 = -(v^2) + c

replace s with a and v with u.

a^2 + u^2 = c

s^2 + v^2 = a^2 + u^2


when s = 0, a = 0

v = u


Where have i gone wrong?
 
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  • #2
F = ms? Force = MassxLength? :confused:

I think something is amiss.
 
  • #3
neutrino said:
F = ms? Force = MassxLength? :confused:

I think something is amiss.

Doesn't have to be a miss, the given force is a function of the displacement.
 
  • #4
radou said:
Doesn't have to be a miss, the given force is a function of the displacement.
Okay, fine. I suppose the constant in some set of units take the value of 1. Sorry about that.
 
  • #5
Sink41 said:
Question copied word for word

1. A particle of mass m moves in a straight line under the action of force F where F = ms, s being the displacement of the particle from O, a fixed point on the line.
When s = -a the velocity of the particle is u. Find the velocity of the particle when s = 0.2. F = ma or F= -ma (i think...)
F= ma. Don't guess- look it up.

a = v(dv/ds)3. If ms = -ma, then s is always equal to -a, so the velocity would always be u, which doesn't look right... Or if I intergate:
Why would the velocity always be u? If s is not 0 then -a is not 0: with non-zero acceleration, the velocity changes.

s = -a

s = -v(dv/ds)

s^2 = -(v^2) + c

replace s with a and v with u.

a^2 + u^2 = c

s^2 + v^2 = a^2 + u^2
You aren't really saying anything here. What is v if not u?
when s = 0, a = 0

v = uWhere have i gone wrong?
You are correct that [itex]v^2+ s^2= C[/itex], a constant.

When s = -a the velocity of the particle is u. Find the velocity of the particle when s = 0.
Is "a" here some given constant? In that case, you have [itex]C= u^2+ a^2[/itex] and so the general formula is [itex]v^2+ s^2= u^2+ a^2[/itex] where u and a are given constants. When s= 0, [itex]v^2= u^2+ a^2[/itex] and so [itex]v= \pm \sqrt{u^2+ a^2}[/itex].
 
  • #6
If F = ma then the particle would speed off in one direction, unless it ended up or started at O with a velocity of zero. If F = -ma then you would get harmonic motion... I thought it might be plus or minus because in the question it didnt say whether the force was towards or away from O.

I was thinking that a was acceleration, which was why i got confused... The velocity can't always be u. In the other questions the teacher used k for any generic constant and used the word acceleration instead of a symbol. I'll just guess that he meant to write k and made a mistake.
 
Last edited:

1. How is force related to acceleration in a mechanics question with variable force?

In mechanics, force is directly proportional to acceleration. This means that as force increases, acceleration also increases, and vice versa. This relationship is described by Newton's second law of motion, which states that the acceleration of an object is equal to the net force acting on it divided by its mass.

2. What is the formula for calculating net force in a mechanics question with variable force?

The formula for calculating net force in a mechanics question with variable force is Fnet = ma, where Fnet is the net force, m is the mass of the object, and a is the acceleration. This formula is derived from Newton's second law of motion.

3. How do you determine the direction of the net force in a mechanics question with variable force?

In a mechanics question with variable force, the direction of the net force is determined by the vector sum of all the individual forces acting on the object. This means that the net force will be in the same direction as the total force acting on the object.

4. Can the net force in a mechanics question with variable force ever be zero?

Yes, the net force in a mechanics question with variable force can be zero. This can occur when all the individual forces acting on an object cancel each other out, resulting in a net force of zero. In this case, the object will either remain at rest or continue moving at a constant velocity.

5. How does the concept of work apply to mechanics questions with variable force?

In mechanics, work is defined as the product of force and displacement. In a mechanics question with variable force, the work done on an object can be calculated by multiplying the net force applied to the object by the distance over which the force is applied. This is represented by the equation W = Fd, where W is work, F is force, and d is displacement.

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