Physics Question

[Homoochan]

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Why would final velocity of an object same whether it is dropped from 10m height or released from top of the incline from the base length 10m? (of course, ignoring air resistance and friction)


Thanks!

 

[TieuBachHo]

Why would final velocity of an object same whether it is dropped from 10m height or released from top of the incline from the base length 10m? (of course, ignoring air resistance and friction)


Thanks!

V(f)^2 = V(o)^2 + 2a delta X. Same initial velocity, same displacement, same acceleration. ==> Same final velocity.

EDIT: That is wrong. It's the conservation of energy!

 

[not respect]

that's not true, the displacement for the incline is longer but the acceleration is weaker. the reason is because of conservation of energy.

 

[TieuBachHo]

that's not true, the displacement for the incline is longer but the acceleration is weaker. the reason is because of conservation of energy.

Yea ... 😀 Thanks. Conservation of energy is right

 

[BerkReviewTeach]

Why would final velocity of an object same whether it is dropped from 10m height or released from top of the incline from the base length 10m? (of course, ignoring air resistance and friction)


Thanks!

Conceptually, you're comparing a system in free fall with a large acceleration (g) traveling a short distance (10m) versus a system travelling down a ramp with a smaller accelaration (g sin theta) traveling a longer distance (10m/sin theta). The two factors balance one another.

From a conservation of energy standpoint, there are two ways to look at it. (1) In both cases (free fall and ramp), mghinitial = 1/2mvexp2.
(2) F x d = F x d which leads to mg x d = mg sin theta x d/sin theta