Can an Object Maintain Constant KE and Experience Net Acceleration?

  • Thread starter Thread starter icanletyougo
  • Start date Start date
  • Tags Tags
    Energy
AI Thread Summary
An object can experience net acceleration while maintaining constant kinetic energy through circular motion, where the direction of velocity changes but its magnitude remains constant. The kinetic energy equation, KE = (0.5)mv^2, indicates that if speed (v) does not change, kinetic energy (KE) also remains constant. In circular motion, centripetal acceleration is present, which alters the direction of the velocity vector without increasing its speed. Thus, while the object accelerates towards the center of the circle, its kinetic energy stays the same. This demonstrates that constant acceleration can occur without a change in kinetic energy.
icanletyougo
Messages
9
Reaction score
0

Homework Statement


Is it possible for an object to experience a net acceleration, yet maintain a constant kinetic
energy?

Homework Equations





The Attempt at a Solution


I think no because the kinetic energy equation is KE = (0.5)mv^2, because KE is not change, so v will not change; therefore, it will have acceleration equal =0.
But I'm not sure.
 
Physics news on Phys.org
What type of motion has a constant acceleration yet never goes any faster?

Hint: Think of something other than straight-line motion.
 
gneill said:
What type of motion has a constant acceleration yet never goes any faster?

Hint: Think of something other than straight-line motion.

circular motion?
 
icanletyougo said:
circular motion?

Okay. What's your argument as to why this is so?
 
gneill said:
What type of motion has a constant acceleration yet never goes any faster?

Will you consider centripetal acceleration constant. It keeps changing directions.
 
ashishsinghal said:
Will you consider centripetal acceleration constant. It keeps changing directions.

True, but it's a constant magnitude and the kinetic energy does not change.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top