Find work done by force given distances and times

In summary,A 1.0 kg body is at rest on a friction-less horizontal air track when a constant horizontal force F acting in the positive direction of an x-axis along the track is applied to the body. A stroboscopic graph of the position of the body as it slides to the right is shown in (the) figure. The force F is applied to the body at t = 0, and the graph records the position of the body at 0.5 second intervals. Between 0.5 and 2.0 seconds, the body does work equal to the change in kinetic energy.
  • #1
dsb1713
4
0

Homework Statement


A 1.0 kg body is at rest on a friction-less horizontal air track when a constant horizontal force F acting in the positive direction of an x-axis along the track is applied to the body. A stroboscopic graph of the position of the body as it slides to the right is shown in (the) figure. The force F is applied to the body at t = 0, and the graph records the position of the body at 0.5 second intervals. How much work is done on the body by the applied force F between t = 0 and t= 2.0 seconds?

At 0.5 seconds, the distance is 0.04 meters, at 1 second it is 0.2 meters, at 1.5 seconds it is 0.44 meter, at 2 seconds it is 0.8 meters.

Homework Equations


Work is equal to the change of kinetic energy. W=Kf-Ki.
Kinetic energy is equal to one half the mass times instantaneous velocity squared. K=1/2 mv^2
Displacement is equal to one half times initial velocity plus final velocity, multiplied by the time. Xf-Xi=(Vi+Vf)/2 *t

The Attempt at a Solution


I am doing this exercise in order to practice for our final test. I've found a solution, but my professor refuses to provide us with an answer key to the practice problems he gave us, so I cannot check whether my work is correct. I've found similar problems online but none with my exact numbers.

Solution attempt:
Because the object starts from rest, I assume the initial kinetic energy will be zero. Thus, because work is the change in kinetic energy, which is Kf minus Ki, it should be equal solely to the final kinetic energy. Therefore, we must find Kf.
Kf=1/2mv^2
We lack the final velocity Vf. However, given that we know the displacement and the time, we can find it by isolating the Vf variable from the following equation: Xf-Xi=(Vi+Vf)/2 *t
Vf will be equal to 0.8m/s
Plugging it into the previous Kinetic energy function, we get a total work of 0.32J.
Is my response correct? Is there a better approach?
Thanks for your time.
 
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  • #2
Hi dsb1713, Welcome to Physics Forums.

I've changed your thread title to: "Find work done by force given distances and times". This is to comply with forum rules that require thread titles to be descriptive of the thread content and the nature of the problem.

What definition do you know for work (other than the change in KE one, which is the work-energy theorem).
 
  • #3
gneill said:
Hi dsb1713, Welcome to Physics Forums.

I've changed your thread title to: "Find work done by force given distances and times". This is to comply with forum rules that require thread titles to be descriptive of the thread content and the nature of the problem.

What definition do you know for work (other than the change in KE one, which is the work-energy theorem).
Thank you for adjusting the title.
I have Work being defined as the product of a Force vector and a Displacement vector. Also, work is equal to the negative change in potential energy.
In this case, I believe we could use the first of these two, using 0.8 meters as our displacement. For the force, using the standard F=m.a equation, I would need to find the acceleration using 2D kinematics formulas.
Am I on the right track?

Thank you for your time.
Edit: Through 2D kinematics I found the acceleration to be 0.4m/s^2. By multiplying it with the mass, I found the force to be 0.4N. Now, by multiplying the force by the distance, I found the Work to be 0.32 Joules, consistent with my previous answer. Thank you so very much!
 
  • #4
Yes, looks like you're on the right track. In fact your initial approach is also good. There are several ways to get to the desired goal.

If this were a real lab where you took measurements you might want to plot a graph of the collected data in a way that you could draw a best fit line to extract the acceleration, then determine the force using f = ma.
 
  • #5
I had the graphics package Kaleidagraph fit the data, and the fit was excellent, with the equation:
$$x=0.18511t^{2.1572}$$
What does that give you for the work at t = 2 sec?

Chet
 
  • #6
Chestermiller said:
I had the graphics package Kaleidagraph fit the data, and the fit was excellent, with the equation:
$$x=0.18511t^{2.1572}$$
What does that give you for the work at t = 2 sec?

Chet
This equation represents x as a function of t, taken from the data provided in the problem?
I'm not certain how could I use this information to help me solve this problem

Thanks for your time
 
  • #7
Hmm. Since we'd expect ##d = \frac{1}{2} a t^2## from the standard kinematics, why not fit 2d vs t2?
 
  • #8
dsb1713 said:
This equation represents x as a function of t, taken from the data provided in the problem?
Yes.
I'm not certain how could I use this information to help me solve this problem
Do you know how to determine the velocity from this equation?
 
  • #9
gneill said:
Hmm. Since we'd expect ##d = \frac{1}{2} a t^2## from the standard kinematics, why not fit 2d vs t2?
It wasn't clear to me from the problem statement that the force F was constant. If it is, then, of course, the fit you recommend is preferred.OOPs. It did say constant. Oh well, Neeevvveeerr mind.

Chet
 
  • #10
dsb1713 said:
This equation represents x as a function of t, taken from the data provided in the problem?
I'm not certain how could I use this information to help me solve this problem

Thanks for your time
Chestermiller said:
Yes.

Do you know how to determine the velocity from this equation?
I assume through derivatives, although this is not in the scope of my current course. Also, yes, I believe the acceleration and force are both constant.
Thanks for your time & help
 

Related to Find work done by force given distances and times

1. What is work done by force?

Work done by force is the amount of energy transferred to an object when a force is applied to it and causes it to move a certain distance in the direction of the force.

2. How is work calculated?

Work is calculated by multiplying the force applied to an object by the distance it moves in the direction of the force. The formula for work is W = F * d, where W is work, F is force, and d is distance.

3. Can work be negative?

Yes, work can be negative. This occurs when the force and the direction of movement are in opposite directions. For example, if you push a box in one direction and it moves in the opposite direction, the work done by your force would be negative.

4. How are units of work measured?

The SI unit of work is joule (J), named after the physicist James Prescott Joule. Other units of work include foot-pound (ft-lb) and calorie (cal).

5. How are time and work related?

Time and work are inversely related. This means that if the same amount of work is done, but the time taken to do it is increased, the amount of force applied will be less. In other words, the longer the time taken to do the work, the less force is required.

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