Body of mass m is subject to a constant force

In summary, The conversation is about solving a problem involving a body of mass m subject to a constant force F1 for a length of time t1, followed by a sudden change to a constant force F2. The equations for v(t) and x(t) are provided, but there is difficulty in accounting for the step function in x(t). The correct analysis involves adding the term F1*t1/m*(t-t1) to account for the change in force during t1-t.
  • #1
jesuslovesu
198
0
[solved] Step Force

Never mind, i think i figured it out

Homework Statement


Suppose a body of mass m is subject to a constant force [tex]F_1[/tex] that acts for a length time [tex]t_1[/tex] and then the force suddenly changes to a constant [tex]F_2[/tex]

Show [tex]v(t) = v_0 + \frac{F_1 t_1}{m} + \frac{F_2(t-t_1 )}{m}[/tex]
[tex]x(t) = x_0 + v_0 t_1 + \frac{F_1 {t_1}^2}{2m} + (v_0 + \frac{F_1 t_1}{m}(t-t_1 ) ) + \frac{F_2}{2m}(t-t_1 )^2 [/tex]

Homework Equations





The Attempt at a Solution



I'm having a hard time with the step force. Here's how I get v(t)

[tex]\int_{v_0}^v dv = \int_0^{t_1} F_1/m dt + \int_{t_1}^{t} F_2/m dt[/tex]
[tex]v(t) = F_1/m t_1 + F_2/m (t - t_1) + v_0[/tex]
Which is fine, but then when it gets to x(t) is where I am having problems:

[tex]\int_{x_0}^{x} dx = \int_{0}^{t_1} F_1/m t_1 dt + \int_{t_1}^{t} F_2/m (t-t_1) dt + \int_0^{t_1} v_0 dt + \int_{t_1}^t v_0 dt[/tex]

Which yields:
[tex]x(t) = x_0 + v_0 t_1 + (v_0)(t-t_1) + F_1/m {t_1}^2 + \frac{F_2 (t-t_1)^2 }{2m}[/tex]

I know that in order to have a step function you need to get rid of [tex]F_1[/tex] after time t_1 because it no longer acts on the object. So I could just add [tex]F_1 t_1 / m (t-t_1)[/tex] but I'm not really sure where the [tex]F_1 t_1/m (t)[/tex] force would come into play.

So my question is: Am I doing the analysis correctly? It seems like taking care of the step function is proving to be a difficultly.
 
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  • #2
Any help would be greatly appreciated. Thanks! Yes, you are doing the analysis correctly. To include the step force, you need to add the following term: F1*t1/m*(t–t1). This term accounts for the change in force during the time period t1-t.
 
  • #3


Dear student,

Your analysis is correct. The step function in this problem can be interpreted as a sudden change in the force acting on the object. This means that the force F_1 acts only for a duration of t_1 and then suddenly changes to F_2. In your calculation for x(t), you have correctly accounted for this change in force by including the term F_1 t_1/m (t-t_1). This term represents the distance traveled by the object due to the force F_1 before it changes to F_2.

In other words, at time t_1, the object has already traveled a distance of v_0 t_1 + F_1 t_1^2/(2m) due to the initial force F_1. After time t_1, the object is then subjected to a new force F_2 and travels a distance of (v_0 + F_1 t_1/m)(t-t_1) + F_2 (t-t_1)^2/(2m). Adding these two distances together gives you the expression for x(t) that you have derived.

I hope this helps clarify your understanding. Keep up the good work!
 

1. What is the equation for calculating the acceleration of an object with mass m under a constant force?

The equation for calculating the acceleration of an object with mass m under a constant force is F = ma, where F is the force applied, m is the mass of the object, and a is the resulting acceleration.

2. How does the magnitude of the constant force affect the acceleration of an object with mass m?

The magnitude of the constant force directly affects the acceleration of an object with mass m. The greater the force applied, the greater the acceleration of the object will be.

3. Does the direction of the constant force have any impact on the acceleration of an object with mass m?

Yes, the direction of the constant force does have an impact on the acceleration of an object with mass m. The force and acceleration will be in the same direction if the force is applied in the same direction as the object's motion, and in opposite directions if the force is applied in the opposite direction of the object's motion.

4. Can an object with mass m experience a constant force and still have a net force of zero?

Yes, an object with mass m can experience a constant force and still have a net force of zero if the force is balanced by an equal and opposite force. In this case, the object will either remain at rest or continue moving with a constant velocity.

5. How can the velocity of an object with mass m be determined using the constant force equation?

The constant force equation can be rearranged to solve for velocity by dividing both sides by m, giving the equation a = F/m. By substituting in the known values for force and mass, the resulting acceleration can then be used to calculate the object's velocity using the equation v = u + at, where u is the initial velocity and t is the time elapsed.

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