Easy question, looking for quick help

  • Thread starter Thread starter IntegrateMe
  • Start date Start date
AI Thread Summary
The weight of an object on the moon is one-sixth of its weight on Earth, but this does not affect its kinetic energy. Kinetic energy is calculated using the formula KE = 1/2 mv², where mass remains constant regardless of location. Therefore, the kinetic energy of a body moving at speed V is the same on both the moon and Earth, leading to a ratio of 1:1. The initial confusion arose from conflating weight with mass in the calculations. Understanding that mass is invariant while weight varies due to gravitational acceleration clarifies the mistake.
IntegrateMe
Messages
214
Reaction score
1
The weight of an object on the moon is one-sixth of its weight on Earth. The ratio of the
kinetic energy of a body on Earth moving with speed V to that of the same body moving with
speed V on the moon is:

A. 6:1
B. 36:1
C. 1:1
D. 1:6
E. 1:36

I did the calculations using KE = 1/2 mv2 and got A, but the answer is C. Any explanations? Thanks.
 
Physics news on Phys.org
Using KE = 1/2 mv^2 you would get the same answer on the moon as on Earth. The formula has MASS in it, not WEIGHT. The mass of an object is the same on the moon as it is on Earth. The weight, F = mg, is different because the gravitation acceleration g is less on the moon.
 
Wow, thanks for that! Such a silly mistake.
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
Back
Top