An object m=2kg, is 10 meters above the ground at the top of a long ramp and starts to roll down?

An object m=2kg is 10 meters above the ground at the top of a long ramp and starts to roll down. If you are ignoring friction, what is the object鈥檚 velocity at the bottom of the ramp? Is this velocity greater than, less than, or equal to the velocity that the ball would have if it were simply dropped straight down to the ground without rolling down the ramp?An object m=2kg, is 10 meters above the ground at the top of a long ramp and starts to roll down?
P.E at the top is converted into K.E. at the bottom


mgh = 0.5mv^2


v =sqrt (2gh) = 14 m/s





In case of the free fall of the object


u = 0


Using , v^2 = u^2 - 2gh


h = sqrt(2gh) = 14 m/s , same as aboveAn object m=2kg, is 10 meters above the ground at the top of a long ramp and starts to roll down?
With no friction to worry about, the object's speed is exactly what it would be if dropped straight down; it just takes longer to get to the bottom than if it fell.





The potential energy here is mgh = (2 kg)(9.81 m/s虏)(10 m) = 196 J.


The resulting kinetic energy Ek is (1/2)mv虏, so v = 鈭?2Ek/m)


= 鈭?(2)(196 J)/(2 kg)) = 14 m/s.
Just an addition to the other answers, which pretty much have it covered, the velocity will actually be in a different direction but the same magnitude.


Velocity is a Vector quantity, therefore direction is Very important.


It seems your teacher set you a sneaky trick question there


the best answer you can give is


The velocity is equal in magnitude but acts in a different direction.