What force gives me this impulse?

In summary, to find the value of Fmax that gives an impulse of 5.6 Ns, you need to integrate the x-component of the force over the time interval from t initial to t final. This will give you the impulse in terms of area, which you can then equate to 5.6 Ns and solve for Fmax.
  • #1
Nightrider55
18
0

Homework Statement


what value of Fmax gives an impulse of 5.6 N\s?

Picture of graph below

http://session.masteringphysics.com/problemAsset/1070437/6/09.EX05.jpg


Homework Equations



Jx=Integral from t initial to t final of Fx(t)dt

^px=Jx


The Attempt at a Solution


I started by setting 5.6 Ns equal to the integral of the area below the curve but I think I am missing a step. I don't fully understand how you would start a problem like this could someone help me to get on the right track, thanks.
 
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  • #2
I think you are on the right track.

You started correctly by setting:

[tex]J_x=\int_{t_i}^{t_f}F_x(t)dt[/tex]

HINT: Next, get the integral in terms of the area.
 
  • #3


I would approach this problem by first understanding the concept of impulse and how it relates to force. Impulse is the product of force and time, and it is a measure of the change in momentum of an object. In this problem, the impulse is given as 5.6 Ns, and we want to find the value of force (Fmax) that would result in this impulse.

To start, we can use the formula J = Ft, where J is the impulse, F is the force, and t is the time interval. In this case, we are given the impulse (J) and we need to find the force (F). Therefore, we can rearrange the formula to solve for F: F = J/t.

Next, we need to determine the time interval (t) over which the impulse is applied. Looking at the graph, we can see that the impulse is represented by the area under the curve. So, we can calculate the time interval by taking the integral of the force (F) with respect to time (t), as shown in the equations above.

Now, we can plug in the given values into the formula F = J/t. The given impulse is 5.6 Ns, and the time interval can be calculated by taking the integral of the force graph. Once we have the time interval, we can solve for Fmax.

In summary, the force (F) that gives an impulse of 5.6 Ns can be found by taking the given impulse and dividing it by the time interval, which can be calculated by taking the integral of the force graph.
 

Related to What force gives me this impulse?

What force gives me this impulse?

The force that gives an object an impulse is known as the "impulsive force". This force is responsible for changing an object's momentum, which is the product of an object's mass and velocity. When an object is acted upon by an impulsive force, its momentum changes, resulting in an impulse.

What is the relationship between force and impulse?

Force and impulse are directly related. An impulse is equal to the average force applied to an object multiplied by the time interval over which the force is applied. In other words, the greater the force applied to an object, the greater the impulse, and the longer the force is applied, the greater the impulse.

How is impulse calculated?

To calculate impulse, you can use the formula I = F * Δt, where I is impulse, F is the average force, and Δt is the time interval over which the force is applied. Impulse is measured in units of Newton-seconds (N*s).

What factors affect the magnitude of an impulse?

The magnitude of an impulse is affected by several factors, including the magnitude of the force applied, the direction of the force, and the duration of the force's application. The greater the force and the longer it is applied, the greater the impulse will be.

How can I increase the impulse on an object?

To increase the impulse on an object, you can either increase the magnitude of the force applied or increase the duration of the force's application. This can be achieved by increasing the mass or velocity of the object, or by applying the force for a longer period of time. However, it is important to note that increasing the impulse on an object may also result in a change in the object's momentum and potentially its direction of motion.

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