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.
 
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...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top