B Why doesn't the gravity vector contribute to KE sliding down a ramp?

AI Thread Summary
The discussion clarifies how to calculate kinetic energy (KE) for a block on a declined ramp. It emphasizes that KE is determined by the equation KE = initial PE - any energy lost due to friction, rather than including work done by gravity separately. This is because including gravity's work would result in double counting, as it is already factored into the change in potential energy (PE). Participants suggest solving the problem using both potential energy and work done by gravity to understand the equivalence of the results. Overall, the conversation enhances comprehension of energy conservation principles in physics.
gazeem
Messages
9
Reaction score
2
TL;DR Summary
For physics questions where, say, a block is at the top of a declined ramp, how come the KE once the block reaches the bottom is determined by:

KE = initial PE - any energy lost due to friction

rather than:

KE = initial PE - any energy lost due to friction + any work done by the vector component of gravity that is parallel to the direction vector of the object.
244340
 
Physics news on Phys.org
gazeem said:
KE = initial PE - any energy lost due to friction + any work done by the vector component of gravity that is parallel to the direction vector of the object
The PE is the same as the work done by the vector component of gravity that is parallel to the direction vector of the object.
 
  • Like
Likes russ_watters, gazeem and Chestermiller
gazeem said:
Summary: For physics questions where, say, a block is at the top of a declined ramp, how come the KE once the block reaches the bottom is determined by:

KE = initial PE - any energy lost due to friction

Or more generally:
KE = (initial PE - final PE) - any energy lost due to friction

gazeem said:
rather than:

KE = initial PE - any energy lost due to friction + any work done by the vector component of gravity that is parallel to the direction vector of the object.
Beacuse that would be double accounting for the work done by gravity, which is already included as:
initial PE - final PE
 
  • Like
Likes gazeem
Ah okay I see. Thank you both so much for your responses, makes more sense now.
 
  • Like
Likes Dale
gazeem said:
Ah okay I see. Thank you both so much for your responses, makes more sense now.

It would be a good exercise to solve the problem both ways: 1) using PE and 2) using W = Fd for the gravitational force down the slope.

This may give you an insight into why physically and mathematically they give the same result.
 
  • Like
Likes gazeem and russ_watters
Consider an extremely long and perfectly calibrated scale. A car with a mass of 1000 kg is placed on it, and the scale registers this weight accurately. Now, suppose the car begins to move, reaching very high speeds. Neglecting air resistance and rolling friction, if the car attains, for example, a velocity of 500 km/h, will the scale still indicate a weight corresponding to 1000 kg, or will the measured value decrease as a result of the motion? In a second scenario, imagine a person with a...
Dear all, in an encounter of an infamous claim by Gerlich and Tscheuschner that the Greenhouse effect is inconsistent with the 2nd law of thermodynamics I came to a simple thought experiment which I wanted to share with you to check my understanding and brush up my knowledge. The thought experiment I tried to calculate through is as follows. I have a sphere (1) with radius ##r##, acting like a black body at a temperature of exactly ##T_1 = 500 K##. With Stefan-Boltzmann you can calculate...
Thread 'Griffith, Electrodynamics, 4th Edition, Example 4.8. (First part)'
I am reading the Griffith, Electrodynamics book, 4th edition, Example 4.8 and stuck at some statements. It's little bit confused. > Example 4.8. Suppose the entire region below the plane ##z=0## in Fig. 4.28 is filled with uniform linear dielectric material of susceptibility ##\chi_e##. Calculate the force on a point charge ##q## situated a distance ##d## above the origin. Solution : The surface bound charge on the ##xy## plane is of opposite sign to ##q##, so the force will be...
Back
Top