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.
Best Answer
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.