According to Tesla, the model S weighs 1961 kg and uses 237.5 watt hours (855000 Joules) to drive one km. However, when I plug this amount of energy and 1000 meters into the formula W=FD, I get the maximum weight that would work to be 87 kg! I am a little confused as to how an almost 2000 kg car can drive 1km with on only 237 watt hours! Here is my work

W=FD 855000 ={X KG x 9.81 m/s2} (1000 m) X=87.15 kg

According to this formula, to drive a 1961 kg car one km, i would need 19237410 Joules, a much larger amount of energy than what Tesla claims it uses! My only guess is that this has to do with an object in motion staying in motion (Newtown's 1st law). I'm very confused. I would appreciate if anyone can shed some light on this for me. My physics professor was stumped on this too. thank you!


You are totally confused in doing this computation. In your computation, you're plugging in the gravity of the Earth. That's not the relevant force acting on the car, which is friction, coming from both the air and the tires. It's easy to see what's going wrong when you draw a force diagram: gravity pulls the car down, but it's counteracted by a normal force. In any case, the car is moving orthogonally to the direction of gravity, so it doesn't need to do any "work" to counter it.

  • $\begingroup$ Hans, I used your points to fix my work. This is what I came up with: W=FD W=(1961 x .67)(1000) W=1313870 Joules or 365 watt hours. Note that .67 is the coefficient of friction between asphalt and rubber. This still does not equal the 237 watt hours. I did not compute air resistance, but I would assume that it would increase the energy needed. $\endgroup$ – Ved. Jun 30 '16 at 22:51
  • $\begingroup$ You're getting in the right direction, but the formula still is not right. You can estimate the friction on the tires as [friction coefficient] x [mass of the car] x [gravity of the Earth], so your answer should contain a factor of $g = 9.81...$ somewhere. See for instance hyperphysics.phy-astr.gsu.edu/hbase/mechanics/frictire.html Air resistance is more complicated and depends on the velocity $v$ of the car. $\endgroup$ – Hans Moleman Jul 1 '16 at 20:04
  • $\begingroup$ I'm confused as to why we would use the gravity of the earth. If I multiply my force (mass x .67) by 9.81, it changes my answer from 365 Wh to 3.578 Kwh. I removed 9.81 from my work because you mentioned in your first post that we are not trying to make the car go against the force of gravity, only friction. $\endgroup$ – Ved. Jul 1 '16 at 22:53
  • $\begingroup$ I looked at the link and saw that the coefficient of friction can change depending on the road conditions (Ie: wet or dry) so I calculated the coefficient of friction using 237.5 wh as my work, 1961 kg as mass, 1000 m =d and a variable, f to be friction. It worked out to to be .44. Could my answer be off by 127 watt hours by normal variation? $\endgroup$ – Ved. Jul 1 '16 at 22:55
  • $\begingroup$ The reason is that the force of friction is proportional to the magnitude of the normal force - but it doesn't go in the same direction. (The normal force goes straight up.) The reason is pretty complicated and you would have to think about what goes on precisely at the surface of the tire. $\endgroup$ – Hans Moleman Jul 1 '16 at 23:56

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