Mechanics - Work and Kinetic Energy question

• thehammer
In summary, the problem involved a 3kg body at rest on a frictionless horizontal air track, with a constant horizontal force F applied in the positive direction of an x-axis. A stroboscope recorded the position of the body at 0.5 s intervals, with the readings as follows: Interval (I) 1: 0.04 m I2: 0.2 m I3: 0.44 m I4: 0.8 m The question asked for the amount of work done on the body by the applied force F between t = 0 and t = 2s. Using the equations K = 1/2mv^2, W = \DeltaK
thehammer

Homework Statement

A 3kg body is at rest on a frictionless 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. The force F is applied to the body at t = 0 when the body is at the origin. A stroboscope records the position of the body at 0.5 s intervals. The readings are as follows.

Interval (I) 1: 0.04 m
I2: 0.2 m
I3: 0.44 m
I4: 0.8 m

How much work is done on the body by the applied force F between t = 0 and t = 2s?

Homework Equations

K = 1/2mv^2
W = $$\Delta$$K
s = ut + 1/2at^2
v^2 = u^2 + 2as

The Attempt at a Solution

I really have no idea why I'm not getting the right answer. I can't see an error in what I've done.

1) Find a
u = 0

a = 2s/t^2
a = 2 x 0.04 / 0.5^2 = 0.32 ms^-2

2) Find v^2

v^2 = ((0 ms^-1)^2) + 2(0.32 ms^-1 x 0.8m)

3) Find the change in kinetic energy and hence the work done

$$\Delta$$K = 1/2mv^2 = 1/2(3kg)(0.512m^2 s^-2) = 0.768 J

---

The answer is meant to be 0.96J. I think I might have done this question months ago but I cannot see where I have gone wrong in the least. Please point out any glaring mistakes :P.

You didn't make any mistakes, but the data seems to be off a bit. (The acceleration is not constant.) Try finding the acceleration using the I4 data.

I did what you suggested and got the question right. Thank you Doc Al. I've a few of questions about this problem.

If the acceleration is not constant then it follows that the force must be variable. This contradicts the first part of the question in which it is stated that the force is constant. Such a situation would require the use of calculus.

In this instance, I did not have to use calculus and got the correct answer. I initially got the answer wrong because I use the first time interval to work out the acceleration, which I assumed was constant. Why does this method work if the acceleration is nonlinear? Is it the average acceleration that is found using this method?

thehammer said:
If the acceleration is not constant then it follows that the force must be variable. This contradicts the first part of the question in which it is stated that the force is constant.
Correct. I think the explanation is simply that whoever made up this problem messed up the data. For example, if the data were:
I1 = 0.05
I2 = 0.2
I3 = 0.45
I4 = 0.8​
then the acceleration would be a constant 0.4 m/s^2.

What textbook is this from?

The question is from "Fundamentals of Physics, 7th Edition," Halliday, Resnick and Walker.

thehammer said:
The question is from "Fundamentals of Physics, 7th Edition," Halliday, Resnick and Walker.
What chapter and problem number?

Chapter 7, question 9. I sincerely hope I didn't write the distance down incorrectly, haha. I'm fairly sure I didn't.

Looks like I don't have that particular edition, but I suspect a typo was involved.

I hope so. This problem was really getting at me earlier when it shouldn't have been because of its simplicity!

1. What is work in mechanics?

Work in mechanics is defined as the force applied to an object multiplied by the distance the object moves in the direction of the force. It is a measure of the energy transfer that occurs when a force is applied to an object and causes it to move.

2. How is work calculated?

Work is calculated by multiplying the force applied to an object by the distance the object 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. What is kinetic energy?

Kinetic energy is the energy an object possesses due to its motion. It is defined as one half of the mass of an object multiplied by the square of its velocity. Kinetic energy is a scalar quantity and is measured in joules (J).

4. How is kinetic energy related to work?

Kinetic energy is related to work through the work-energy theorem, which states that the net work done on an object is equal to the change in its kinetic energy. This means that when work is done on an object, its kinetic energy changes and vice versa.

5. What is the difference between work and power?

Work and power are both measurements of energy, but they represent different aspects of energy. Work measures the energy transfer that occurs when a force is applied to an object, while power measures the rate at which work is done. In other words, power is the amount of work done per unit of time.

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