A ball of mass $0.37 \,\mathrm{kg}$ is thrown upward along the vertical with a initial speed of $14 \,\mathrm{m\cdot s^{-1}}$, and reaches a maximum height of $8.4 \mathrm{m}$.

  • What is the work done by air resistance on the ball?
  • Assuming that the air resistance performs roughly the same job during the descent, calculate the speed of the ball when it returns to the starting point.

How do I calculate the work of the air resistance?

  • 1
    $\begingroup$ Hi Damian. Welcome to Physics.SE. Our homework policy states that, homework questions that doesn't show any individual work effort by the author but, asks other users to solve it instead - should not be allowed. Please have in mind that you read our policy before asking such questions :-) $\endgroup$ – Waffle's Crazy Peanut Jan 9 '13 at 15:40

Well, the energy must be conserved, so the energy lost while it goes up must be the work performed by air resistance. If you have an initial velocity and a maximum altitude, then you can calculate the difference of energies: $E_{start}-E_{end}$.Being the starting energy only kinetic energy, and the end energy only potential energy.

  • $\begingroup$ I have calculated the kinetic energy and the potetial energy; the latter is less than the first; so the work done by the air is W = Ek - Ep ? $\endgroup$ – DamianFox Jan 9 '13 at 15:02
  • $\begingroup$ YES! You first have an energy $E_k$, then you launch it and the air starts to make the system lose energy by friction, that is the work that the air does. Works are energies, in this case they're energies that you take out or put in the system. As the universe total energy must be conserved, then that work must be the difference of the energies the system has at the beginning and the ending. $\endgroup$ – MyUserIsThis Jan 9 '13 at 15:06
  • $\begingroup$ Thanks! The speed of the ball when it returns to the starting point can be calculated using the work found $$ W = 1/2*m*v^2 $$ Right? $\endgroup$ – DamianFox Jan 9 '13 at 15:17
  • $\begingroup$ The speed of the ball when it comes down again you can calculate doing the same. The problem says that roughly the same amount of energy will be disipated (which doesn't look realistic btw), then, just to the potential energy when it's up, do $E_p-W$ and that will be the kinetic energy when it comes down again, so: $E_p-W=\frac{1}{2}mv^2$ $\endgroup$ – MyUserIsThis Jan 9 '13 at 16:37

From the initial conditions you could compute the maximum altitude without air friction.

$$s(t) = .5gt^2 + v_{start} t$$ $$v(t)= 0:\, \rightarrow t_{max-alt} \rightarrow s(t_{max-alt})$$

then you compare this altitude to the altitude given 8.4m in your problem. this will give you the difference in potential energy. from there you should be able to recover the work done by air resistance

  • $\begingroup$ This solution is much more complicated than the one proposed by @Ferfer93 $\endgroup$ – Bernhard Jan 9 '13 at 15:40
  • $\begingroup$ 'Complicated', yes, but it also gives you an insight into how much altitude you lost because of air resistance. it is always a trade-off between simplicity and information,if you use energy conservation in mechanics instead of the equations of motion. $\endgroup$ – elcojon Jan 9 '13 at 15:45
  • $\begingroup$ I would always choose the approach with the least amount of work to obtain the information you need. Especially with homework question, you should never give an answer to a question that isn't asked. $\endgroup$ – Bernhard Jan 9 '13 at 16:16

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