The discussion revolves around the energy dissipated when braking a car, specifically a 1200-kg car coming to a stop from an initial speed of 30 m/s. Participants are examining the calculations related to kinetic energy and the implications of energy loss during braking.
The discussion is active, with various interpretations being explored regarding the energy calculations and the correct representation of units. Some participants provide guidance on the importance of balancing equations and considering the signs of energy values, while others raise questions about the assumptions made in the original calculation.
There is a focus on ensuring that the left and right sides of equations are balanced, including units. Participants also note the lack of explicit information regarding the direction of energy change in the context of the problem.
physkid1 said:1. How much energy is dissipated in braking a 1200-kg car to a stop from an initial speed of 30 m/s?
2. 1/2 mv(final)^2 - 1/2 mv(initial)^2
3. 1/2 x 1200 x 0^2 - 1/2 x 1200 x 30^2 = 540000J
is this right ? cheers
Shouldn't the right hand side be -540000 ?physkid1 said:1/2 x 1200 x 0^2 - 1/2 x 1200 x 30^2 = 540000J
omoplata said:Shouldn't the right hand side be -540000 ?
And where did the J come from? If you don't have units in the left side of the equation, why did you put units on the right side?
Scalars can be negative.PeterO said:Energy is a scalar, so is never negative [nor positive, nor up nor down nor North or any other direction] it is just a value.
PeterO said:The unit of energy is Joule, so a J is probably very appropriate.
omoplata said:Scalars can be negative.
Example 1: If the potential energy due to a point mass infinity at an infinite distance is assumed to be zero, then the potential energy at a finite distance from the mass is negative.
Example 2: If the increase of kinetic energy in the car in this problem is negative, that means the energy has decreased. i.e. the car has lost energy.
Also refer to https://www.physicsforums.com/showthread.php?t=270612" thread.
The equation can be written down so the units on the left hand side and right hand side are equal, which the OP has not done. Also, the OP ignores the sign of the answer. So, technically, it is not an "equation" at all.
Getting into the habit of writing down correct equations to begin with can help students a lot, in avoiding common errors that happen because of misinterpretation of units etc.
The potential energy change is approximately 10 kg * 9.81 m/s^2 * 12m.PeterO said:OK try this:
By how much does the Gravitational Potential Energy of a 10 kg mass change for a trip between ground level and the top of a 12 m building?
omoplata said:The potential energy change is approximately 10 kg * 9.81 m/s^2 * 12m.
What's your point?
PeterO said:I was interested to see if you fill a calculation up with units as you go - you apparently do; never seen anyone else do that - and I was also interested to see if you would first want to know whether the mass was going from the ground to the building, or the building to the ground - which didn't worry you.
Very simple - yes, by the calculation there should be a negative sign on the other side of the equation. However, that is the energy removed from the car. Thus, the energy dissipated by the brakes is the positive value of that.omoplata said:Shouldn't the right hand side be -540000 ?
And where did the J come from? If you don't have units in the left side of the equation, why did you put units on the right side?