A ball of mass $0.37\text{ kg}$ is thrown upward along the vertical with a initial speed of $14\text{ m/s}$, and reaches a maximum height of $8.4\text{ m}$.

a) What is the work done by air resistance on the ball?

I came up with

$$F \times\text{Distance}(8.4) = .5 \times\text{Weight}(.37)\times(0-14^2)$$

But my answer doesn't match with the book's answer which is $-5.8\text{ J}$.

Not sure if I"m doing this wrong; if there's some special way to calculate air resistance that isn't the formula $W = .5(\text{Mass})(v_f^2-v_i^2)$, with $v_f$ being the final speed and $v_i$ being initial speed.

Part B = Assume that air resistance does about the same work on the downward trip. Estimate the speed of the ball as it returns to its starting point.

  • $\begingroup$ What exactly is your answer? I see you wrote out the formula you used to find your answer, but you never included the actual answer. $\endgroup$ – David Z Jul 7 '13 at 22:55
  • $\begingroup$ Oh the actual answer is -5.8 J, but I think I figured out how to get it, by adding Kinetic Energy Plus Potential Energy. I had no idea what that meant, since I have never taken a physics course, but seemed relatively simple. $\endgroup$ – user26761 Jul 7 '13 at 23:11
  • $\begingroup$ You also have to keep in mind the work done by gravity. $\endgroup$ – fibonatic Jul 7 '13 at 23:11
  • $\begingroup$ Yeah I didn't think about that either (9.8) until I looked what "Potential Energy was". Now I have to somehow estimate the speed of the ball as it returns to its starting point. $\endgroup$ – user26761 Jul 7 '13 at 23:13

Look at your formula.

$$W = .5(\text{Mass})(v_f^2 - v_i^2)$$

What part of it says something about the air? If you had done the same experiment underwater, presumably the water would stop the ball more effectively than air does. How would your formula look different? How have you used the given information about the height the ball traveled?

You haven't. The formula can't be complete because it doesn't tell you anything about the air. It's just a formula for work in general. It doesn't involve the height (which it shouldn't if it's just a general formula, because work is a change in kinetic energy and kinetic energy doesn't depend on height).

Before you use a formula to solve a problem, you need to look at each part of that formula and see how it relates to the different parts of the problem.

The formula you've written is related to the formula for kinetic energy

$$KE = .5(\text{Mass})v^2$$

Can you see how? Can you see that the formula is for the change in the kinetic energy?

So if work is the change in kinetic energy, where does that energy go? Two places: some of it goes into the gravitational potential energy, since the ball is rising. Some of it goes into pushing the air around.

If you want just the part that corresponds to pushing the air around, you can take the total work and subtract that part that went into the gravitational energy.

  • $\begingroup$ hmmmm Yeah I understand the Kinetic portion , but not the potential portion. Although I can elect to memorize the potential formula until I conceptualize it when I actually take physics. I do NOT understand how to estimate the SPEED however as it returns to its starting point, assuming air resistance is the same. $\endgroup$ – user26761 Jul 7 '13 at 23:24
  • $\begingroup$ Ohh nevermind I get it, I can just plug the WORK back into the formula: W = 1/2 * m * v^2 $\endgroup$ – user26761 Jul 7 '13 at 23:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.